What I think you mean is (for example), something like this $$\frac{\partial}{\partial x}\frac{\partial}{\partial y}w$$
This is usually denoted by $$\frac{\partial^2 w}{\partial x\partial y}$$and is defined by $$\lim_{\delta x\to 0}\lim_{\delta y \to 0}\left(\frac{f(x+\delta x,y+\delta y,z)-f(x+\delta x,y,z)-f(x,y+\delta y,z)+f(x,y,z)}{\delta x\delta y}\right)$$
As an example, if we let $f(x,y,z)=x^2y^3$, then $$\frac{\partial^2 f}{\partial x\partial y}=\frac{\partial }{\partial x}\frac{\partial }{\partial y}(x^2y^3)=\frac{\partial }{\partial x}(3x^2y^2)=6xy^2$$
$$f(x(z,w),y(z,w),z,w)=g(z,w)$$
Don't confuse $\frac{\partial f}{\partial z}$ with $\frac{\partial g}{\partial z}$
$\frac{\partial f}{\partial z}$ means the partial derivative of $f$ with respect to $z$ for $x$ independant of $z$, which is not the same for $g$.
Thus the correct expression is :
$$\frac{\partial g}{\partial z} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial z}+ \frac{\partial f}{\partial y}\frac{\partial y}{\partial z}+ \frac{\partial f}{\partial z}$$
It is clear that $$\frac{\partial g}{\partial z} \neq \frac{\partial f}{\partial z}$$
However, loosely the same symbol $f$ is used instead of two distinct symbols $f$ and $g$. This is acceptable for people familiar with this ambiguous writting. They are aware of apparent contradictions which can arise and they avoid mistakes in mentally making the distinction between the two different signification of the symbol $f$.
IN ADDITION :
Since this seems difficult to well understand, consider a concret example :
$$f(x,y,z,w)=5x+4y+3z+2w$$
$$\frac{\partial f}{\partial x}=5\quad;\quad \frac{\partial f}{\partial y}=4\quad;\quad \frac{\partial f}{\partial z}=3$$
Then consider another function $g(z,w)$ defined from the preceeding function in the particular case of
$$x(z,w)=6z+7w\quad\text{and}\quad y(z,w)=8z+9w$$
$$\frac{\partial x}{\partial z}=6\quad;\quad \frac{\partial y}{\partial z}=8$$
$$g(z,w)=5(6z+7w)+4(8z+9w)+3z+2w$$
$$g(z,w)=65z+73w$$
Obviously $5x+4y+3z+2w$ is something else than $65z+73w$ . So $f(x,y,z,w)$ and $g(z,w)$ are not a same function. They cannot be confused.
$$\frac{\partial g}{\partial z}=65$$
We see that $\frac{\partial g}{\partial z}$ is obtained directly by partial differentiation of $g(z,w)$. But instead of, if we want to compute $\frac{\partial g}{\partial z}$ indirectly from the properties of the functions $f(x,y,z,w)$ , $x(z,w)$ and $y(z,w)$ we apply the chain rule of derivations :
$$\frac{\partial g}{\partial z} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial z}+ \frac{\partial f}{\partial y}\frac{\partial y}{\partial z}+ \frac{\partial f}{\partial z}$$
$$\frac{\partial g}{\partial z} =5*6+4*8+3=65$$
As expected, the same result as before $65$ is obtained without expressing explicitely $g(z,w)$. This is purely a consequence of the chain rule. There is no need for more explanation insofar the theory of the chain rule was studied and understood.
Best Answer
The quantity $\frac{\partial f(x,y)}{\partial x}$ is defined as the derivative of $f$ with respect to its first argument. Even though you will eventually assign $y = \phi(x)$, the quantity $\frac{\partial f(x,y)}{\partial x}$ still refers to the (perhaps hypothetical) assessment of "how much does $f$ vary when $x$ is varied and $y$ (its second argument) is held constant." Meanwhile, the total derivative $\frac{df(x,y)}{dx}$ assesses all ways that $f$ varies with $x$, including any "indirect" contribution through how $y$ may depend on $x$ (i.e. $\phi$). Generally speaking, $$ \frac{df(x,y)}{dt} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial t} $$
Letting $x=t$ we have, \begin{align} \frac{df(x,y)}{dx} &= \frac{\partial f}{\partial x} \frac{\partial x}{\partial x} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial x}\\ &= \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial x} \end{align}
Letting $y = \phi(x)$ we have, $$ \frac{df(x,y)}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \phi'(x) $$
Notice that, $$ \phi'(x)=0\ \implies\ \frac{df}{dx}=\frac{\partial f}{\partial x} $$
The partial derivative under question is "how $f$ varies while we vary $x$ and hold its other arguments constant", $\frac{\partial f}{\partial x}$, which still makes sense even if $y=\phi(x)$.
In the case of $f(x,y) = e^{xy}$ we have, $$ \frac{\partial f(x,y)}{\partial x} = ye^{xy} $$ and \begin{align} \frac{df(x,y)}{dx} &= ye^{xy} + xe^{xy}\phi'(x)\\ &= e^{xy}\big{(}y + x\phi'(x)\big{)} \end{align}
which is, as you have found and we would hope, equivalent to thinking of $f$ as a function of one variable $x$ to find by "product rule", $$ \frac{df(x)}{dx} = e^{x\phi(x)}\big{(}\phi(x) + x\phi'(x)\big{)} $$