Chain rule problem without giving f(x) The questions is
If $f(u,v,w)$ is differentiable and $u=x-y$, $v=y-z$, and $w=z-x$, show that
$$\begin{equation}
\frac{\partial{f}}{\partial{x}}+\frac{\partial{f}}{\partial{y}}+\frac{\partial{f}}{\partial{z}}=0
\end{equation}$$
The things that I could only get is $$\begin{equation}
\frac{\partial{f}}{\partial{x}}+\frac{\partial{f}}{\partial{y}}+\frac{\partial{f}}{\partial{z}}=\frac{\partial{f}}{\partial{u}}+\frac{\partial{f}}{\partial{v}}+\frac{\partial{f}}{\partial{w}}
\end{equation}$$
At that point, I even don`t know what $f(x)$ is....so how can I get the answer? 

It doesn't matter what $f$ is. When you want partial derivative with respect to à variable, you need to take every path that leads to this variable. Since $u$ and $w$ depends on $x$,
$$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial w}\frac{\partial w}{\partial x}=\frac{\partial f}{\partial u}-\frac{\partial f}{\partial w}$$
Same goes for $y$ and $z$
$$\frac{\partial f}{\partial y}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial y}=-\frac{\partial f}{\partial u}+\frac{\partial f}{\partial v}$$
$$\frac{\partial f}{\partial z}=\frac{\partial f}{\partial v}\frac{\partial u}{\partial z}+\frac{\partial f}{\partial w}\frac{\partial w}{\partial z}=-\frac{\partial f}{\partial v}+\frac{\partial f}{\partial w}$$
If you sum those three equations, you find the $0$ you are looking for.
