Finding partial derivative given $x,y,z$ related by $yx = \ln(x+z)$ 
We are given that $x, y, z$ are related by the equation $yx = \ln(x + z)$.
What is the partial derivative of $z$ with respect to $x$ at the point $(\frac e2, \frac 2e, \frac e2)$?

I'm sorry I don't have any work on this problem; I'm just really confused where to start because I just started learning partial derivatives.
 A: Well, first we need to isolate $z$:
\begin{align}yx&=\ln(x+z)\\
e^{yx}&=e^{\ln(x+z)}\\
e^{yx}&=x+z\tag{$\dagger$}\\
z&=e^{yx}-x\end{align}
Now we compute the partial derivate with respect to $x$ (treating $y$ as a constant): \begin{align}\frac{\partial z}{\partial x}&=\frac{\partial}{\partial x}\left(e^{yx}-x\right)\\
&=ye^{yx}-1\end{align}
Finally we compute the value of this at $(x,y,z)=(\frac e2, \frac 2e, \frac e2)$:
\begin{align}ye^{yx}-1&=\frac 2e\times e^{\frac 2e\times\frac e2}-1\\
&=\frac 2e\times e^1-1\\
&=\frac{2e}{e}-1\\
&=2-1\\
&=1\end{align}

$(\dagger)$ Note that $e^{\ln(a)}=a$
A: Another way using implicit differentiation.
Deriving both sides of $yx=\ln(x+z)$    with respect to $x$ we have:
$$
x\frac{\partial y}{\partial x}+y=\frac{1}{x+z}\left( \frac{\partial z}{\partial x} +1\right)
$$
So:
$$
1+\frac{\partial z}{\partial x}=\left(x\frac{\partial y}{\partial x}+y \right)(x+z)
$$
and, since $(x+z)=e^{xy}$,
$$
\frac{\partial z}{\partial x}=\left(x\frac{\partial y}{\partial x}+y \right)e^{xy}-1
$$
If $y$ is independent from $x$ than $\frac{\partial y}{\partial x}=0$ and the derivative at $(\frac e2, \frac 2e, \frac e2)$ is
$$
\frac{\partial z}{\partial x}=ye^{xy}-1=\frac 2e \cdot e -1=1
$$
A: Express $z$ as a function of $x$ and $y$ and calculate the derivata of $z$ as a function of $x$, as if $y$ was a constant.
