Chain Rule in Partial Derivative

$\displaystyle f\left(u,v,w\right)$ is differentiable and $\displaystyle u=\;x-y,\; v=\;y-z,\;w=\;z-x$, show that $\displaystyle \frac{\partial{f}}{\partial{x}}\;+\;\frac{\partial {f}}{\partial{y}}\;+\;\frac{\partial{f}}{\partial{ z}}\;=\;0$.

I think i'll use chain rule and that equation will be $\displaystyle \frac{\partial{f}}{\partial{u}}\frac{\partial{u}}{ \partial{x}}\;+\;\frac{\partial{f}}{\partial{v}}\f rac{\partial{v}}{\partial{y}}\;+\;\frac{\partial{f }}{\partial{w}}\frac{\partial{w}}{\partial{z}}\;=\ ;0$

From now on i can not find any value for $\displaystyle \frac{\partial{f}}{\partial{u}}\;$ ,, $\displaystyle \frac{\partial{f}}{\partial{v}}\;$ and $\displaystyle \;\frac{\partial{f}}{\partial{x}}$. (Headbang) So how can i find these values?

If that's not the way of solution, i need some hints.