Partial derivative of a two variables function, one of which dependent on the other I found this exercise on the book of multivariable calculus from which I'm studying:
"Find the partial derivative $\frac{\partial{z}}{\partial{x}}$ and the total derivative $\frac{\text{d}z}{\text{d}x}$ of $z(x,y)=e^{xy}$ where $y=\phi(x)$."
Now, this to me looks like a function of a single variable $f:\mathbb{R}\to\mathbb{R}$ and so in this case the partial derivative of $f$ with respect to $x$ and total derivative would be equivalent; in particular, I end up with something like:
$$\frac{\text{d}z}{\text{d}x}=e^{xy}(\phi(x)+x\phi'(x))$$
In the solution, while the result for the total derivative is the same as mine, the partial derivative of $f$ with respect to $x$ is written as follows:
$$\frac{\partial{z}}{\partial{x}}=ye^{xy}$$
Why is this the case?
Since the partial derivative of $f$ with respect to $x$ shows the incremental behaviour of the function as $x$ changes, shouldn't I account for the presence of $x$ in the functional representation of $y$ while computing the derivative with respect to $x$?
Sorry in advance for the super basic question :)
 A: 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",
$$
\frac{df(x)}{dx} = e^{x\phi(x)}\big{(}\phi(x) + x\phi'(x)\big{)}
$$
