Let, $f$ be a differentiable function on $\mathbb{R}$ and $g(x,y) = f(xy)$, show that
$x\frac{\partial g}{\partial x } -y\frac{\partial g}{\partial y} = 0$
This is the chapter on the multivariable chain rule. I have tried doing the following $G(x,y) = xy$ and hence $g = f \circ G $. By the chain rule, we get
$dg = \begin{pmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \end{pmatrix}$ $\begin{pmatrix} y \\x \end{pmatrix}$
But, how do I go about proving the above relation? I'm fairly confused by the presentation of the multivariable chain rule in my book and I think I have severe misunderstanding of what is even going on. The lack of good examples in the book do not help the matter at all. Any help here would be greatly appreciated.
Just use the chain rule $$ \frac{\partial g}{\partial x } = y\cdot f' $$ where $f'$ is the derivative of the function $f$. Similar relation holds for $\frac{\partial g}{\partial y}$. Applying these to your expression we get: $$ x\frac{\partial g}{\partial x } -y\frac{\partial g}{\partial y} =x\cdot f'\cdot y - y\cdot f' \cdot x = 0 $$