I am given that $z=f(x^2+y^2)$ and need to show that $y\frac{\partial z}{\partial x}-x\frac{\partial z}{\partial y}=0$.
Listed below is my attempt. Is this correct? I feel that it seems too simplistic.
$\large{ \frac{\partial z}{\partial x} = f'(x^2+y^2) \dot{} 2x\\ \frac{\partial z}{\partial y} = f'(x^2+y^2) \dot{} 2y }$
Hence,
$$ \begin{align*} y\frac{\partial z}{\partial x}-x\frac{\partial z}{\partial y} &= f'(x^2+y^2) \dot{} 2x \times y - f'(x^2+y^2) \dot{} 2y \times x \\&=2xy \times f'(x^2+y^2) -2xy \times f'(x^2+y^2) \\&=0 \end{align*} $$
Your calculations are correct, you just need to make sure that $f$ is differentiable.
– Git Gud Mar 17 '13 at 16:35