Answer
$\displaystyle \frac{\partial z}{\partial x}=\frac{yz}{e^{z}-xy},\qquad\frac{\partial z}{\partial y}=\frac{xz}{e^{z}-xy}$
Work Step by Step
$e^{z}=xyz \displaystyle \qquad/\frac{\partial}{\partial x}$, (y is constant, z is a function of x)
$\displaystyle \frac{\partial}{\partial x}(e^{z})=\frac{\partial}{\partial x}(xyz)$
... RHS: product rule
$e^{z}=\displaystyle \frac{\partial z}{\partial x}=y(x\frac{\partial z}{\partial x}+z\cdot 1)$
$e^{z}\displaystyle \frac{\partial z}{\partial x}-xy \displaystyle \frac{\partial z}{\partial x}=yz$
$(e^{z}-xy)\displaystyle \frac{\partial z}{\partial x}=yz$,
$\displaystyle \frac{\partial z}{\partial x}=\frac{yz}{e^{z}-xy}$.
$e^{z}=xyz \displaystyle \qquad/\frac{\partial}{\partial y}$, (x is constant, z is a function of y)
$\displaystyle \frac{\partial}{\partial y}(e^{z})=\frac{\partial}{\partial y}(xyz)$
... RHS: product rule
$e^{z}\displaystyle \frac{\partial z}{\partial y}=x(y\frac{\partial z}{\partial x}+z\cdot 1)$
$e^{z}\displaystyle \frac{\partial z}{\partial y}-xy\frac{\partial z}{\partial y}=xz$
$(e^{z}-xy)\displaystyle \frac{\partial z}{\partial y}=xz$
$\displaystyle \frac{\partial z}{\partial y}=\frac{xz}{e^{z}-xy}$
