Answer
$\frac{\partial z}{\partial y}=\frac{xz}{e^z -xy}$
$\frac{\partial z}{\partial x}=\frac{yz}{e^z -xy}$
Work Step by Step
The equation $e^z=xyz$ is given. To find its partial derivatives, one must differentiate both sides of the equation with respect to either $x$ or $y$ while maintaining that $z$ is an implicit function of either $x$ or $y$.
To find the partial derivative $f_x$:
$\frac{\partial}{\partial x} {e^z} = \frac{\partial}{\partial x} {xyz}$
${e^z} \frac{\partial z}{\partial x}= \frac{\partial z}{\partial x}{xy}+yz$
Seperating the partial derivative:
$\frac{\partial z}{\partial x}=\frac{yz}{e^z -xy}$
To find the partial derivative $f_x$:
$\frac{\partial}{\partial y} {e^z} = \frac{\partial}{\partial y} {xyz}$
${e^z} \frac{\partial z}{\partial x}= \frac{\partial z}{\partial y}{xy}+yxz$
Seperating the partial derivative:
$\frac{\partial z}{\partial x}=\frac{yz}{e^z -xy}$
