Answer
\[
\begin{aligned}
\frac{\partial w}{z} &=-\frac{z}{w} \\
\frac{\partial w}{\partial x} &=-\frac{x}{w} \\
\frac{\partial w}{y} &=-\frac{y}{w}
\end{aligned}
\]
Work Step by Step
We are given that
\[
4=\left(w^{2}+x^{2}+y^{2}+z^{2}\right)^{3 / 2}
\]
Thus, w is a function of three variable, so differentiating it with respect to $x$:
\[
\begin{aligned}
\frac{\partial}{\partial x}\left(\left(x^{2}+y^{2}+z^{2}+w^{2}\right)^{3 / 2}\right) &=\frac{d}{d x}(4) \\
\frac{3}{2} \sqrt{x^{2}+y^{2}+z^{2}+w^{2}} *\left(2 x+2 w \frac{\partial w}{\partial x}\right) &=0 \quad \because \sqrt{x^{2}+y^{2}+z^{2}+w^{2}}=\text { positive number } \\
2 x+2 w \frac{\partial w}{\partial x} &=0 \\
\frac{\partial w}{\partial x} &=-\frac{x}{w}
\end{aligned}
\]
Differentiating it with respect to y:
\[
\begin{aligned}
\frac{\partial}{\partial y}\left(\left(x^{2}+y^{2}+z^{2}+w^{2}\right)^{3 / 2}\right) &=\frac{d}{d y}(1) \\
\frac{3}{2} \sqrt{x^{2}+y^{2}+z^{2}+w^{2}} *\left(2 y+2 w \frac{\partial w}{\partial y}\right) &=0 \quad \because \sqrt{x^{2}+y^{2}+z^{2}+w^{2}}=\text { positive number } \\
2 y+2 w \frac{\partial w}{\partial y} &=0 \\
-\frac{y}{w}=\frac{\partial w}{\partial x} &
\end{aligned}
\]
Differentiating it with respect to z:
\[
\begin{aligned}
\frac{d}{d z}(1)=\frac{\partial}{\partial z}\left(\left(w^{2}+x^{2}+y^{2}+z^{2}\right)^{3 / 2}\right) & \\
\frac{3}{2} \sqrt{w^{2}+x^{2}+y^{2}+z^{2}} *\left(2 y+2 w \frac{\partial w}{\partial z}\right) &=0 \quad \because \sqrt{w^{2}+x^{2}+y^{2}+z^{2}}=\text { positive number } \\
0=2 z+2 w \frac{\partial w}{\partial z} & \\
-\frac{z}{w}=\frac{\partial w}{\partial z} &
\end{aligned}
\]
