Answer
$ \frac{\partial z }{\partial x}=\frac{x}{(1-z)}$
$\frac{\partial z }{\partial y}=\frac{y}{(z -1)}$
Work Step by Step
$$
x^{2}-y^{2}+z^{2}-2z=4
$$
To find $\frac{\partial z}{\partial x}$, we differentiate implicitly with respect to x, being careful to treat $y$ as a constant:
$$
\begin{aligned}
\frac{\partial }{\partial x} (x^{2}-y^{2}+z^{2}-2z) &=\frac{\partial }{\partial x} (4)\\
\frac{\partial }{\partial x} (x^{2})-\frac{\partial }{\partial x} ( y^{2}) + \frac{\partial }{\partial x} (z^{2} )-2\frac{\partial }{\partial x} (z )&=\frac{\partial }{\partial x} (1) \\
2x -(0)+(2z \frac{\partial z }{\partial x})) -2\frac{\partial z }{\partial x}&=(0)
\\
2x +2z \frac{\partial z }{\partial x} -2\frac{\partial z }{\partial x}&=0
\\
2x +2\frac{\partial z }{\partial x} (z -1)&=0
\\
2\frac{\partial z }{\partial x} (z -1)&=-2x
\\
\frac{\partial z }{\partial x} &=\frac{-x}{(z -1)}\\
&=\frac{x}{(1-z)}.
\end{aligned}
$$
To find $\frac{\partial z}{\partial y}$, we differentiate implicitly with respect to $y$, being careful to treat $x$ as a constant:
$$
\begin{aligned}
\frac{\partial }{\partial y} (x^{2}-y^{2}+z^{2}-2z) &=\frac{\partial }{\partial y} (4)\\
\frac{\partial }{\partial y} (x^{2})-\frac{\partial }{\partial y} ( y^{2}) + \frac{\partial }{\partial y} (z^{2} )-2\frac{\partial }{\partial y} (z )&=\frac{\partial }{\partial y} (4) \\
2(0) -2(y)+(2z \frac{\partial z }{\partial y})) -2\frac{\partial z }{\partial y}&=(0)
\\
-2y +2z \frac{\partial z }{\partial y} -2\frac{\partial z }{\partial y}&=0
\\
-2y +2\frac{\partial z }{\partial y} (z -1)&=0
\\
2\frac{\partial z }{\partial y} (z -1)&=2y
\\
\frac{\partial z }{\partial y} &=\frac{y}{(z -1)}\\
\end{aligned}
$$
