Problem with chain rule

Question

$z=x^2*f(\frac{y}{x^2})$, $f$ is differentiable.
Show that $z$ satisfies the equation $x\frac{\partial z}{\partial x}+2y\frac{\partial z}{\partial y}=2z$.

How should i use the chain rule over this?

Answer 1

Recall the partial chain rule: $$\frac{\partial U}{\partial x}=\frac{\partial U}{\partial z}\frac{\partial z}{\partial x}$$. So what is $$\frac{\partial z}{\partial x}$$ and $$\frac{\partial z}{\partial y}$$, if we apply these to $$z=x^2 f(y/x^2) \implies \frac{\partial z}{\partial x}=\frac{\partial}{\partial x}(x^2)f(y/x^2)+x^2\frac{\partial}{\partial x}f(y/x^2)\frac{\partial }{\partial x}(y/x^2)= x^2 \frac{\partial}{\partial x}f(u)*\frac{\partial}{\partial x}(u)+2x f(u)=2x f(u) +\frac{-2 x^2 y}{x^3}\frac{\partial}{\partial x}f(u)=2x(\frac{-y}{x^2}\frac{\partial}{\partial x}f(u)+f(u))=2x((-u)\frac{\partial}{\partial x}f(u)+f(u)$$ where $u=u(x,y)=\frac{y}{x^2}$

Now repeat with $\dfrac{\partial}{\partial y}$