Answer
\[
-\frac{x}{z}=\frac{\partial z}{\partial x}
\]
\[
-\frac{y}{z}=\frac{\partial z}{\partial y}
\]
Work Step by Step
We are given that
\[
1=\left(z^{2}+y^{2}+x^{2}\right)^{3 / 2}
\]
Differentiate the variable $ z $ in relation to $ x $ with $ y $ constant
\[
\begin{aligned}
\frac{\partial}{\partial x}\left(\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}\right) &=\frac{d}{d x}(1) \\
\frac{3}{2} \sqrt{x^{2}+y^{2}+z^{2}} *\left(2 x+2 z \frac{\partial z}{\partial x}\right) &=0 \quad \because \sqrt{x^{2}+y^{2}+z^{2}}=1 \\
2 x+2 z \frac{\partial z}{\partial x} &=0 \\
-\frac{x}{z}=\frac{\partial z}{\partial x} &
\end{aligned}
\]
And:
\[
\begin{aligned}
\frac{d}{d y}(1)=\frac{\partial}{\partial y}\left(\left(z^{2}+y^{2}+x^{2}\right)^{3 / 2}\right) & \\
0=\frac{3}{2} \sqrt{x^{2}+y^{2}+z^{2}} *\left(2 y+2 z \frac{\partial z}{\partial y}\right) & \quad \because1= \sqrt{z^{2}+y^{2}+x^{2}} \\
0=2 y+2 z \frac{\partial z}{\partial y} & \\
-\frac{y}{z}=\frac{\partial z}{\partial y} &
\end{aligned}
\]
