Answer
$z_x=\dfrac{yz}{e^z-xy}$ and $z_y=\dfrac{xz}{e^z-xy}$
Work Step by Step
Recall Equation 7: $z_x=-\dfrac{F_x}{F_z}$ and $z_y=-\dfrac{F_y}{F_z}$
Given: $e^z=xyz$
This gives: $e^z-xyz=0$
Let us consider that $F(x,y,z)=e^z-xyz=0$
Here, we have $F_x=-yz\\F_y=-xz\\F_z=e^z-xy$
Equation (1) becomes: $z_x=-\dfrac{-yz}{e^z-xy}$ and $z_y=-\dfrac{-xz}{e^z-xy}$
Therefore, $z_x=\dfrac{yz}{e^z-xy}$ and $z_y=\dfrac{xz}{e^z-xy}$
