Given $$\phi(x,y)=e^yf\left(y\exp\left(\frac{x^2}{2y^2}\right)\right)$$ Prove that: $$\left(x^2-y^2\right)\frac{ \partial\phi }{\partial x}+xy \frac{ \partial\phi }{\partial y}=xy\,\phi(x,y)$$
I tried to use the chain rule , but it became kind of messy and I can't tell where I went wrong or even whether I'm doing it wrong.
Thanks.
That statement cannot possible be correct without some assumption concerning $f$, since the LHS of the equality that you wish to prove depends upon $f'$, whereas the RHS depends only upon $f$. In fact, the equality that you want to prove is$$\frac{x e^{\frac{x^2}{2 y^2}+y} \left(x^2+x y-y^2\right) f'\left(y e^{\frac{x^2}{2 y^2}}\right)}y=xye^yf\left(y e^{\frac{x^2}{2 y^2}}\right).$$Take, say, $f=1$, and this is clearly false.