Answer
$z_x=\dfrac{x}{1-z}$ and $z_y=\dfrac{y}{z-1}$
Work Step by Step
We are given that $x^2-y^2+z^2-2z=4$
Re-arrange as: $x^2-y^2+z^2-2z-4=0$
Consider $F(x,y,z)=x^2-y^2+z^2-2z-4$
$F_x=2x$ and $F_y=-2y$ and $F_z=2z-2$
Use Equation 7 which is: $z_x=-\dfrac{F_x}{F_z}$ and $z_y=-\dfrac{F_y}{F_z}$
$z_x=-\dfrac{F_x}{F_z}=-\dfrac{2x}{2z-2}=\dfrac{x}{1-z}$ and $z_y=-\dfrac{F_y}{F_z}=-\dfrac{2y}{2z-2}=\dfrac{y}{z-1}$
Hence, we have $z_x=\dfrac{x}{1-z}$ and $z_y=\dfrac{y}{z-1}$
