Why that rotation does what it does
It's really probably better to think of multiplication as making rotations, although of course division and multiplication are really the same. In that spirit, look at this:
$$
\frac{z}{b-ai}=z(\frac{b+ai}{a^2+b^2})
$$
Here we can see that $z$ is being multiplied by a complex number of unit length (here you can see it's already normalized) on the right hand side, which we know effects a rotation on the complex plane. It isn't normalized , so it will contribute dilation, but that's OK: two steps later the inverse transformation contributes a dilation factor that cancels this one's dilation out. All that matters is that this transformation carries our line onto the $x$-axis.
So on the step you're at, the line has been translated so that it goes through the origin, so its points all fall on $ax+by=0$.
Let $(x',y')$ be on $ax+by=0$. After rotation by the above mapping, $(x',y')\mapsto (bx'-ay',ax'+by')/(a^2+b^2)$. But since $ax'+by'=0$, this is just $((bx'-ay')/(a^2+b^2),0)$, so the line has been laid on the $x$-axis, as advertised.
From the last paragraph you can see how this can be reverse engineered. If one had applied an arbitrary rotation $(x,y)\mapsto((cx-dy)/(c^2+d^2),cx+dy)$, and they were armed with the equation $ax+by=0$, and you asked them "What choice of $c$ and $d$ makes this rotation put $x,y$ on the $x$ axis? They might say "how about $c=a$ and $d=b$?"
Moved this here since it turned out to be a digression.
Looking at the map as an $\Bbb R$ linear transformation on $\Bbb C$, and viewing the inputs as column vectors $[x,y]^\top$ that correspond to $x+yi$, you can confirm that this matrix accomplishes the transformation by multiplication on the left of the column vectors:
$$
\frac{1}{a^2+b^2}\begin{bmatrix}b&-a\\a&b\end{bmatrix}
$$
It's a matrix with determinant $1$ (of course it should be, since it represents a complex number of unit modulus) and so it's a rotation of $\Bbb R^2$ (which is being identified with $\Bbb C$.)
"variable" is a dangerous word to throw around.
In mathematics, strictly speaking we don't need to name the arguments of the functions, and a function has only one argument. Every function is a function from a set $A$ to a set $B$ and this description doesnt talk about variables.
Now in some cases (all the time in physics) we want to talk about functions of several variables, and more importantly, we want to do change of variables, without changing the name of the function.
For example, if you drop a rock into a well and you consider its speed $s$, altitude $z$, and acceleration $a$, any of those quantities is technically a function of any of the other, so we want to write things like $s(z), a(z), a(s), \ldots$ but where the two $a$ denotes, mathematically, different functions.
When dealing with functions with multiple variables, things get worse when you consider partial derivatives. For example if temperature $T$ of a gas is a function of volume $V$ and pression $P$, but the size and the pression of the container changes with time $t$, you may consider the $T$ as a function of $t$ and $P$.
Now the funny thing is that $dT/dP$ isn't the same thing wether you are talking about the function $T(P,V)$ or $T(P,t)$.
Back to your question, functions from $f : \Bbb C$ to $\Bbb C$ are often talked about as functions $\Bbb R^2$ to $\Bbb C$ with two real arguments $x,y$.
When you do a real change of variable like $x' = (x+y)/\sqrt 2, y' = (x-y)/\sqrt 2$, the two new variables have nothing to do with each other and in this example the change of variable is a mere rotation of the plane.
Say we want to make the $\Bbb C$-linear change of variables $z = x+iy, \bar z = x-iy$ (or in the other direction $x = (z+\bar z)/2, y = i(\bar z-z)/2$) ($z$ and $\bar z$ are just name)s.
Here $z$ takes complex values and $\bar z$ is the conjugate of $z$ when $x$ and $y$ are real. Note that for the change of variable to make sense, we are plugging something syntaxically complex ($i(z-\bar z)/2$) into $y$.
So maybe everything makes better sense if instead of saying that $x$ and $y$ are real variables, we consider them as complex variables (we will note this function $\hat f_1 : \Bbb C^2 \to \Bbb C$, with arguments $x$ and $y$). Because then, the change of variables is almost like a "complex rotation" (we call the resulting function $\hat f_2 : \Bbb C^2 \to \Bbb C$, with arguments $z$ and $\bar z$).
And this time, $z$ and $\bar z$ are two complex variables of $\hat f_2$ that have nothing to do with each other, just like $x$ and $y$ for $\hat f_1$. It just happens that $\bar z$ is the conjugate of $z$ exactly when $x$ and $y$ are real.
You can then talk about $d\hat f_2/dz$ and $d\hat f_2/d\bar z$ and the following are equivalent :
1) the original function $f: \Bbb C \to \Bbb C$ is holomorphic
2) $d\hat f_2/d\bar z = 0$
3) $\hat f_2(z,\bar z)$ only depends on $z$ and not on $\bar z$
4) $i d\hat f_1/dx = d\hat f_1/dy$
5) $\hat f_1(x,y)$ only depends on the value of $x+iy$ and not on the particular (possibly complex) values of $x$ and $y$.
Best Answer
The nomenclature of $\dfrac{\partial}{\partial z}$ and $\dfrac{\partial}{\partial\bar{z}}$ is confusing because it gives the impression that these are really partial derivatives with respect to two independent variables, $z$ and $\bar{z}$. However, it is clear that $z$ and $\bar{z}$ are not independent.
Differentiable Functions and Conformal Maps
A differentiable function on $\mathbb{R}$ locally looks like a linear function, that is, there is a real constant, called $f'(x)$, so that for small $h$, $$ f(x+h)=f(x)+f'(x)h+o(h)\tag{1} $$ Analogously, a differentiable function on $\mathbb{C}$ satisfies $(1)$ for some complex number $f'(x)$.
Multiplication on $\mathbb{C}$ acts as a rotation and radial scale when viewed as an action on $\mathbb{R}^2$. Thus, if $f$ is differentiable on $\mathbb{C}$, $$ f(z+h)-f(z)=f'(z)h+o(h)\tag{2} $$ That is, when $h$ is small, $h\mapsto f(z+h)-f(z)$ looks like a scaled rotation. For this reason, a differentiable function on $\mathbb{C}$ is called conformal: small features are replicated (scaled and rotated) and angles are preserved.
Complex Conjugation and Orientation Reversal
Complex conjugation, $z\mapsto\bar{z}$, is an orientation reversing isometry. Thus, when composed with a conformal map, either before or after, the composition is an orientation-reversing conformal map. Furthermore, double composition yields an orientation-preserving conformal map; for example, if $f(z)$ is conformal, then so is $\overline{f(\bar{z})}$.
As a function on $\mathbb{R}^2$, complex conjugation can be represented by the matrix $\begin{bmatrix}1&0\\0&-1\end{bmatrix}$.
Conformal and Conjugate Conformal
The partial derivatives of a general differentiable function on $\mathbb{R}^2$ given by $x+iy\mapsto u+iv$ are usually given in a $2\times2$ Jacobian matrix: $$ \frac{\partial(u,v)}{\partial(x,y)}=\begin{bmatrix}\frac{\partial u}{\partial x}&\frac{\partial v}{\partial x}\\\frac{\partial u}{\partial y}&\frac{\partial v}{\partial y}\end{bmatrix}\tag{3} $$ The Cauchy-Riemann equations specify that $\dfrac{\partial u}{\partial x}=\dfrac{\partial v}{\partial y}$ and $\dfrac{\partial u}{\partial y}=-\dfrac{\partial v}{\partial x}$, which agrees with the following basis for the orientation-preserving conformal Jacobians on $\mathbb{R}^2$: $$ \left\{\begin{bmatrix}1&0\\0&1\end{bmatrix},\begin{bmatrix}0&1\\-1&0\end{bmatrix}\right\}\tag{4} $$ Note that the determinant of any linear combination of these matrices has positive determinant (thus orientation is preserved).
The following basis for the orientation-reversing conformal Jacobians on $\mathbb{R}^2$ follows by composing conjugation with $(4)$: $$ \left\{\begin{bmatrix}1&0\\0&-1\end{bmatrix},\begin{bmatrix}0&1\\1&0\end{bmatrix}\right\}\tag{5} $$ Note that the determinant of any linear combination of these matrices has negative determinant (thus orientation is reversed).
Using $(4)$ and $(5)$, we can break any Jacobian into conformal and conjugate conformal parts. Using the component-wise orthogonality that exists among the bases, we can write the conformal part as $$ \frac{1}{2}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)\begin{bmatrix}1&0\\0&1\end{bmatrix}+\frac{1}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)\begin{bmatrix}0&1\\-1&0\end{bmatrix}\tag{6} $$ and the conjugate conformal part as $$ \frac{1}{2}\left(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right)\begin{bmatrix}1&0\\0&-1\end{bmatrix}+\frac{1}{2}\left(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\right)\begin{bmatrix}0&1\\1&0\end{bmatrix}\tag{7} $$ $\dfrac{\partial}{\partial z}$, $\dfrac{\partial}{\partial\bar{z}}$, and Quaternions
The definitions of $\dfrac{\partial}{\partial z}$ and $\dfrac{\partial}{\partial\bar{z}}$ say $$ \begin{align} \frac{\partial}{\partial z}(u+iv) &=\frac{1}{2}\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)(u+iv)\\ &=\frac{1}{2}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)+\frac{i}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)\tag{8} \end{align} $$ and $$ \begin{align} \frac{\partial}{\partial\bar{z}}(u+iv) &=\frac{1}{2}\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)(u+iv)\\ &=\frac{1}{2}\left(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right)+\frac{i}{2}\left(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\right)\tag{9} \end{align} $$ The space of $2\times2$ Jacobians has $4$ dimensions, so trying to represent these $4$ dimensions with the $2$ dimensions of $\mathbb{C}$, using $\dfrac{\partial}{\partial z}$ and $\dfrac{\partial}{\partial\bar{z}}$, obscures something.
There is a common matrix representation of the complex numbers as $2\times2$ real matrices where $$ \begin{align} \mathbf{1}&\leftrightarrow\begin{bmatrix}1&0\\0&1\end{bmatrix}\tag{10}\\ \mathbf{i}&\leftrightarrow\begin{bmatrix}0&1\\-1&0\end{bmatrix}\tag{11} \end{align} $$ However, there is also a matrix representation of the quaternions as $2\times2$ complex matrices where, in addition to $(10)$ and $(11)$, $$ \begin{align} \mathbf{j}&\leftrightarrow\begin{bmatrix}i&0\\0&-i\end{bmatrix}\tag{12}\\ \mathbf{k}&\leftrightarrow\begin{bmatrix}0&-i\\-i&0\end{bmatrix}\tag{13} \end{align} $$ Embed $(8)$ and $(9)$ in the quaternions to get $$ \left(\frac{\partial}{\partial z}(u+iv)\right)\mathbf{1} =\frac{\mathbf{1}}{2}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)+\frac{\mathbf{i}}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)\tag{14} $$ and $$ \left(\frac{\partial}{\partial\bar{z}}(u+iv)\right)\mathbf{j} =\frac{\mathbf{j}}{2}\left(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right)+\frac{\mathbf{k}}{2}\left(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\right)\tag{15} $$ Finally, substituting $(10)$-$(13)$ into $(14)$ and $(15)$, it becomes apparent, upon comparison with $(6)$ and $(7)$, that $\left(\dfrac{\partial}{\partial z}(u+iv)\right)\mathbf{1}$ represents the conformal part of the Jacobian and $\left(\dfrac{\partial}{\partial\bar{z}}(u+iv)\right)\mathbf{j}$ represents the conjugate conformal part.
Conclusion
For a general $f:\mathbb{C}\mapsto\mathbb{C}$, $\dfrac{\partial}{\partial z}f$ can be mapped to the conformal part of the $2\times2$ Jacobian, $\dfrac{\partial f}{\partial z}=\dfrac{\partial(u,v)}{\partial(x,y)}$, and $\dfrac{\partial}{\partial\bar{z}}f$ can be mapped to the conjugate conformal part. It is merely convenience of notation that we write $\dfrac{\partial}{\partial z}$ and $\dfrac{\partial}{\partial\bar{z}}$ because $\dfrac{\partial}{\partial z}f\;\mathrm{d}z+\dfrac{\partial}{\partial\bar{z}}f\;\mathrm{d}\bar{z}=\mathrm{d}f$. However, they are not true partial derivatives, but $2$ pieces of a $2\times2$ Jacobian composed of $4$ partial derivatives.
So, to answer the question asked, $z\mapsto\bar{z}$ is conjugate conformal, so $\frac{\partial}{\partial z}\bar{z}=0$; therefore, $\bar{z}$ acts like a constant under $\frac{\partial}{\partial z}$.