Answer
\[
0=-x \frac{\partial z}{\partial y}+y \frac{\partial z}{\partial x}
\]
Work Step by Step
Let $f$ be a differential function of one variable:
\[
\left(y^{2}+x^{2}\right)f=z
\]
Let $u=y^{2}+x^{2}, \quad f(u)=z$
\[
\begin{array}{l}
\frac{\partial z}{\partial x}=\frac{d z}{d u} \frac{\partial u}{\partial x} \\
\frac{\partial z}{\partial y}=\frac{d z}{d u} \frac{\partial u}{\partial y}
\end{array}
\]
Let's consider $\frac{\partial u}{\partial y}=2 y$ and $\frac{\partial u}{\partial x}=2 x,$
\[
\frac{\partial z}{\partial y}=2 y f^{\prime}(u)
\]
\[
\frac{\partial z}{\partial x}=2 x f^{\prime}(u)
\]
Thus, here we can write
\[
\begin{array}{l}
y \frac{\partial z}{\partial x}-x \frac{\partial z}{\partial y}=2 x y f^{\prime}(u)-2 x y f^{\prime}(u) \\
0=-x \frac{\partial z}{\partial y}+y \frac{\partial z}{\partial x}
\end{array}
\]
