Write
$$f(x,y)=u(x,y)+i v(x,y),\quad g(\xi,\eta)=a(\xi,\eta)+ib(\xi,\eta)$$
where $u$, $v$, $a$, $b$ are realvalued functions defined in $A$, resp. $B$. Then by definition of $g$ one has
$$a(\xi,\eta)+ib(\xi,\eta)=g(\xi+i\eta)=\overline{f(\xi-i\eta)}=u(\xi,-\eta)-iv(\xi,-\eta)$$
and therefore
$$a(\xi,\eta)=u(\xi,-\eta),\quad b(\xi,\eta)=-v(\xi,-\eta)\qquad\bigl((\xi,\eta)\in B\bigr)\ .$$
It follows that, e.g. $$b_\eta(\xi,\eta)=-v_y(\xi,-\eta)\cdot(-1)=v_y(\xi,-\eta)\ .$$
Since $u$ and $v$ satisfy the CR-equations in the variables $x$ and $y$ we conclude that
$$a_\xi(\xi,\eta)=u_x(\xi,-\eta)=v_y(\xi,-\eta)=b_\eta(\xi,\eta)\ ,$$
and similarly
$$a_\eta(\xi,\eta)=-u_y(\xi,-\eta)=v_x(\xi,-\eta)=-b_\xi(\xi,\eta)\ .$$
This shows that $g$ fulfills the CR-equations in the variables $\xi$ and $\eta$.
But there is also a direct approach, which in my view is simpler and more in tune with a complex world description.
As $f$ is holomorphic in $A$, for each point $z_0\in A$ (held fixed in the following) there is a complex number $C$ such that
$$f(z)-f(z_0)=C(z-z_0)+o(|z-z_0|)\qquad (z\in A, \ z\to z_0)\ .$$
Let a point $w_0\in B$ be given, and put $z_0:=\bar w_0$. Then by definition of $g$ one has
$$g(w)-g(w_0)=\overline{f(\bar w)}-\overline{f(\bar w_0)}=\overline{f(\bar w)-f( z_0)}=\overline{C(\bar w -z_0)+o(|\bar w-z_0|)}\qquad(w\in B)\ .$$
As $|\bar w -z_0|=|w-w_0|$ it follows that
$$g(w)-g(w_0)=\bar C(w-w_0)+o(|w-w_0|)\qquad(w\in B, \ w\to w_0)\ .$$
It follows that $g'(w_0)=\bar C$, and as $w_0\in B$ was arbitrary, we conclude that $g$ is holomorphic in $B$.
A good rule of thumb for complex analysis is that, if something has gone wrong, $0$ is almost certainly to blame.
Here is the difference quotient for $\overline{z}$ at $0$ taken along two different paths in the complex plane:
Tending to $0$ along the positive imaginary axis produces:
$\lim_{h\rightarrow0}\frac{\overline{\left(0+ih\right)}-\overline{0}}{ih}=\lim_{h\rightarrow0}\frac{-ih}{ih}=-1$
On the other hand, tending to $0$ along the positive real axis produces:
$\lim_{h\rightarrow0}\frac{\overline{\left(0+h\right)}-\overline{0}}{h}=\lim_{h\rightarrow0}\frac{h}{h}=1$
which is completely different. Since the limits do not agree, the limit of this difference quotient as $h$ tends to $0$ in $\mathbb{C}$ does not exist. Thus, the conjugate function is not complex differentiable at $0$.
Generally, when working "from first principles" as it were, one has to rely on the two-dimensionality of $\mathbb{C}$ (along with its algebraic/"arithmetic" properties) to show that things are different than the case of single real-variable analysis. Indeed, it is exactly because the condition of holomorphy requires these two-dimensional limits to exist that holomorphic functions are so marvelously well-behaved (in stark contrast to functions of multiple real variables).
Best Answer
The Wirtinger derivatives are defined by $$\frac{\partial}{\partial \overline z} = \frac{1}{2} \left(\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}\right) \quad \text{and} \quad \frac{\partial}{\partial z} = \frac{1}{2} \left(\frac{\partial}{\partial x} - i \frac{\partial}{\partial y}\right)$$
Let $\psi = u + iv$ where $u = \operatorname{Re}\psi$ and $v = \operatorname{Im}\psi$.
Then \begin{align} \frac{\partial\overline\psi}{\partial z} &= \frac{1}{2} \left(\frac{\partial}{\partial x} - i \frac{\partial}{\partial y}\right)(u-iv) \\ &= \frac{1}{2}\left[(u_x - v_y) - i(u_y + v_x)\right] \end{align} Comparing real and imaginary parts, we see that $\frac{\partial\overline\psi}{\partial z} = 0$ if and only if $u_x = v_y$ and $u_y = -v_x$.
But since $\psi$ is holomorphic it satisfies the Cauchy-Riemann equations, so this is true.