Answer
\[
0=-\frac{\partial z}{\partial y}+2 \frac{\partial z}{\partial x}
\]
Work Step by Step
Let $f$ be a differential function of one variable and
\[
(2 y+x)f=z
\]
Let $x+2 y=u, \quad f(u)=z$
\[
\begin{array}{l}
\frac{\partial z}{\partial x}=\frac{d z}{d u} \frac{\partial u}{\partial x} \\
\frac{\partial z}{\partial y}=\frac{d z}{d u} \frac{\partial u}{\partial y}
\end{array}
\]
Consider $\frac{\partial u}{\partial y}=2$ and $\frac{\partial u}{\partial x}=1$
\[
\frac{\partial z}{\partial x}=(1) f^{\prime}(u)
\]
\[
\frac{\partial z}{\partial y}=(2) f^{\prime}(u)
\]
So we can write
\[
\begin{array}{l}
2 \frac{\partial z}{\partial x}-\frac{\partial z}{\partial y}=2 f^{\prime}(u)-2 f^{\prime}(u) \\
2 \frac{\partial z}{\partial x}-\frac{\partial z}{\partial y}=0
\end{array}
\]
