Answer
$z_x=\dfrac{yz}{e^z-xy}$ and $z_y=\dfrac{xz}{e^z-xy}$
Work Step by Step
We are given that $e^z=xyz$
Re-arrange as: $e^z-xyz=0$
Consider $F(x,y,z)=e^z-xyz=0$
$F_x=-yz$ and $F_y=-xz$ and $F_z=e^z-xy$
Use Equation 7: $z_x=-\dfrac{F_x}{F_z}$ and $z_y=-\dfrac{F_y}{F_z}$
$z_x=-\dfrac{F_x}{F_z}=-\dfrac{-yz}{e^z-xy}=\dfrac{yz}{e^z-xy}$ and $z_y=-\dfrac{F_y}{F_z}=-\dfrac{-xz}{e^z-xy}=\dfrac{xz}{e^z-xy}$
Hence, we have $z_x=\dfrac{yz}{e^z-xy}$ and $z_y=\dfrac{xz}{e^z-xy}$
