Show that $x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y} = 0$?

Question

Given $z = f(\frac{(x+y)}{(x-y)})$ for $f$ a $C^1$ function, I need to show $x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y} = 0$. I tried setting $u=\frac{x+y}{x-y}$ but eneded up with $xf'(u)\frac{(x-y)-(x+y)}{(x-y)^2} + yf'(u)\frac{(x-y)+(x+y)}{(x-y)^2}$ which doesn't seem to simplify to 0, at least the way I did it. Am I approaching it the wrong way?