Prove $(y^2-zx)\frac{\partial u}{\partial x}+(x^2-yz)\frac{\partial u}{\partial y}+(z^2-xy)\frac{\partial u}{\partial z}=0$, $u=f(x^2+2yz,y^2+2zx)$

Question

If $u=f(x^2+2yz,y^2+2zx)$, prove that $$ (y^2-zx) \frac{\partial u}{\partial x} + (x^2-yz)\frac{\partial u}{\partial y}+(z^2-xy)\frac{\partial u}{\partial z}=0 $$

I am trying to prove the above but I am stuck. Could anyone help?