Answer
$\displaystyle \frac{\partial z}{\partial x}=\frac{x}{1-z},\qquad\frac{\partial z}{\partial y}=\frac{y}{z-1}$.
Work Step by Step
$x^{2}-y^{2}+z^{2}-2z =4 \displaystyle \qquad/\frac{\partial}{\partial x}$, (y is constant, z is a function of x)
$\displaystyle \frac{\partial}{\partial x}(x^{2}-y^{2}+z^{2}-2z) =\displaystyle \frac{\partial}{\partial x}(4)$
$2x-0+2z \displaystyle \frac{\partial z}{\partial x}-2\frac{\partial z}{\partial x}=0$
$(2z -2)\displaystyle \frac{\partial z}{\partial x}=-2x$
$\displaystyle \frac{\partial z}{\partial x}=\frac{-2x}{2z-2}$
$\displaystyle \frac{\partial z}{\partial x}=\frac{-2(x)}{-2(-z+1)}$
$\displaystyle \frac{\partial z}{\partial x}=\frac{x}{1-z}$
$x^{2}-y^{2}+z^{2}-2z =4 \displaystyle \qquad/\frac{\partial}{\partial y}$, (x is constant, z is a function of y)
$\displaystyle \frac{\partial}{\partial y}(x^{2}-y^{2}+z^{2}-2z)=\frac{\partial}{\partial y}(4)$
$0-2y +2z\displaystyle \frac{\partial z}{\partial y}-2\frac{\partial z}{\partial y}=0$
$(2z -2)\displaystyle \frac{\partial z}{\partial y}=2y$
$\displaystyle \frac{\partial z}{\partial y}=\frac{2y}{2z-2}$
$\displaystyle \frac{\partial z}{\partial y}=\frac{y}{z-1}$.
