Partial Derivative Chain rule proof Show that if $f$ is a function of the variables x and y (independent variables), and the latter are changed to independent variables u and v where $u = e^{y/x}$ and $x = x^2+y^2$, then
$x\frac{\partial{f}}{\partial{x}} + y\frac{\partial{f}}{\partial{y}} = 2v\frac{\partial{f}}{\partial{v}} $
I have no idea to start, I know how chain rule works for partial derivates when there the intermediate variables u and v are in terms of only one independent variable but I don't know what do to when it is in terms of two.
Can someone show me cause I have been stuck on this question for at least an hour.
Thank you so much!
 A: Consider the function $f(x,y) = g(u,v) = g(e^{y/x}, x^2 + y^2)$.
Note also that $\frac{\partial u}{\partial x} = -\frac{ye^{y/x}}{x^2} = -\frac{yu}{x^2}$ and $\frac{\partial u}{\partial y} = \frac{e^{y/x}}{x} = \frac{u}{x}$ and $\frac{\partial v}{\partial x}= 2x$ and $\frac{\partial v}{\partial y}= 2y$.
Then 
$$
\frac{\partial f}{\partial x} = \frac{\partial g}{\partial x} = \frac{\partial g}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial g}{\partial v}\frac{\partial v}{\partial x} = -\frac{yu}{x^2}\frac{\partial g}{\partial u} + 2x\frac{\partial g}{\partial v}
$$
$$
\frac{\partial f}{\partial y} = \frac{\partial g}{\partial y} = \frac{\partial g}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial g}{\partial v}\frac{\partial v}{\partial y} = \frac{u}{x}\frac{\partial g}{\partial u} + 2y\frac{\partial g}{\partial v}
$$
So that
$$
x\frac{\partial f}{\partial x} = -\frac{yu}{x}\frac{\partial g}{\partial u} + 2x^2\frac{\partial g}{\partial v}
$$
$$
y\frac{\partial f}{\partial y} = \frac{yu}{x}\frac{\partial g}{\partial u} + 2y^2\frac{\partial g}{\partial v}
$$
Putting it together, knowing that $\frac{\partial f}{\partial v} = \frac{\partial g}{\partial v}$,  we get the desired result:
$$
x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y} = 2x^2\frac{\partial g}{\partial v} + 2y^2\frac{\partial g}{\partial v} = 2(x^2 + y^2)\frac{\partial g}{\partial v} = 2v\frac{\partial f}{\partial v}
$$
