Answer
\[
\begin{array}{l}
0=\frac{\partial w}{\partial z}+\frac{\partial w}{\partial y}+\frac{\partial w}{\partial x}
\end{array}
\]
Work Step by Step
We are given that
\[
(x-y, y-z, z-x)f=w
\]
Let $-y+x=u, \quad -z+y=v \quad$ and $\quad -x+z=t$
\[
\begin{array}{l}
\frac{\partial w}{\partial x}=\frac{\partial w}{\partial u} \frac{\partial u}{\partial x}+\frac{\partial u}{\partial v} \frac{\partial v}{\partial x}+\frac{\partial u}{\partial t} \frac{\partial t}{\partial x} \\
\frac{\partial w}{\partial y}=\frac{\partial w}{\partial u} \frac{\partial u}{\partial y}+\frac{\partial w}{\partial v} \frac{\partial v}{\partial y}+\frac{\partial w}{\partial t} \frac{\partial t}{\partial y} \\
\frac{\partial w}{\partial z}=\frac{\partial w}{\partial u} \frac{\partial u}{\partial z}+\frac{\partial w}{\partial v} \frac{\partial v}{\partial z}+\frac{\partial w}{\partial t} \frac{\partial t}{\partial z}
\end{array}
\]
Let's consider
\[
1=\frac{\partial u}{\partial x}, \quad \frac{\partial u}{\partial y}=-1
\]
\[
\begin{array}{l}
\frac{\partial v}{\partial y}=1, \quad \frac{\partial v}{\partial z}=-1 \\
\frac{\partial t}{\partial z}=1, \quad \frac{\partial t}{\partial x}=-1
\end{array}
\]
Thus, here we can write
\[
\begin{array}{l}
\frac{\partial w}{\partial x}+\frac{\partial w}{\partial y}+\frac{\partial w}{\partial z}=\left[\frac{\partial w}{\partial u}(1)+0+\frac{\partial w}{\partial t}(-1)\right]+\left[\frac{\partial w}{\partial u}(-1)+\frac{\partial u}{\partial v}(1)+0\right]+\left[0+\frac{\partial w}{\partial v}(-1)+\frac{\partial w}{\partial t}(1)\right] \\
\frac{\partial w}{\partial x}+\frac{\partial w}{\partial y}+\frac{\partial w}{\partial z}=\frac{\partial w}{\partial u}-\frac{\partial w}{\partial u}+\frac{\partial w}{\partial v}-\frac{\partial w}{\partial v}+\frac{\partial w}{\partial t}-\frac{\partial w}{\partial t} \\
0=\frac{\partial w}{\partial z}+\frac{\partial w}{\partial y}+\frac{\partial w}{\partial x}
\end{array}
\]
