Krdytkyu

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 $f:mathbb{R}tomathbb{R}$ 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:

$$frac{text{d}z}{text{d}x}=e^{xy}(phi(x)+xphi'(x))$$

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 :)

It is possible for $phi(x)$ to be stationary with respect to $x$ wiggles while $f(x,phi(x))$ is not stationary due to its remaining explicit dependence on $x$. When we write the partial derivative, we by-definition require that $y$ is constant, or equivalently that $phi(x)$ is constant while we wiggle x, so it doesn't contribute to the variation (hence why it gets called "partial"). The "total" variation of $f$ makes no such restriction, and thus depends on the general wiggle of both x and $phi(x)$.

I assume you have an intuition / understanding of the following equality,
$$
frac{df(x,y)}{dt} = frac{partial f}{partial x} frac{partial x}{partial t} + frac{partial f}{partial y} frac{partial y}{partial t}
$$

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 $y$ 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,
$$
frac{partial f}{partial x} = ye^{xy}
$$
and
begin{align}
frac{df(x,y)}{dx} &= ye^{xy} + xe^{xy}phi'(x)\
&= e^{xy}big{(}y + xphi'(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",
$$
frac{df(x)}{dx} = e^{xphi(x)}big{(}phi(x) + xphi'(x)big{)}
$$