Chapter 14 - Partial Derivatives - 14.5 The Chain Rule - 14.5 Exercises - Page 984: 33

$z_x=\dfrac{yz}{e^z-xy} \\ z_y=\dfrac{xz}{e^z-xy}$

We know that $z_x=-\dfrac{F_x}{F_z}$ and $z_y=-\dfrac{F_y}{F_z}$ Now, we will re-write the given equation as: $e^z-xyz=0$ Consider $F(x,y,z)=e^z-xyz=0$ $F_x=-yz; F_y=-xz; F_z=e^z-xy$ $$z_x=-\dfrac{F_x}{F_z}=-\dfrac{-yz}{e^z-xy}=\dfrac{yz}{e^z-xy} \\ z_y=-\dfrac{F_y}{F_z}=-\dfrac{-xz}{e^z-xy}=\dfrac{xz}{e^z-xy}$$