How to prove this differential equation? $$z=\frac{y}{f((x^2)+(y^2))}$$
$$\frac{1}{x}\frac{\partial z}{\partial x} + \frac{1}{y}\frac{\partial z}{\partial y}=\frac{z}{y^2}.$$
This looks obvious but spent already a lot of time on it. 
Any Hint or idea?
 A: The derivative of $z= \frac{y}{f(x,y)}$ with respect to x is $\frac{\partial z}{\partial x}= \frac{-y\frac{\partial f}{\partial x}}{f^2}$ and the derivative of $z= \frac{y}{f(x,y)}$ with respect to y is $\frac{\partial z}{\partial y}= \frac{f(x,y)- y\frac{\partial f}{\partial y}}{f^2}$.  
Further, since $f(x, y)= f(x^2+ y^2)$, $\frac{\partial f}{\partial x}= 2x f'$ and $\frac{\partial f}{\partial y}= 2y f'$ where f' is the derivative of f with respect to the single variable.
A: $$\frac{1}{x}\frac{\partial z}{\partial x} + \frac{1}{y}\frac{\partial z}{\partial y}=\frac{z}{y^2}.$$
Note that $ f'=\frac {df}{d(x^2+y^2)}$ for simplicity
$$\frac{\partial f}{\partial x}=2xf' \,;\, \frac{\partial f}{\partial y}=2yf'$$
$$\frac{1}{x}\left (\frac{-2xyf'}{f^2}\right ) + \frac{1}{y}\frac{\partial z}{\partial y}=\frac{1}{fy}.$$
$$\frac{1}{x}\left (\frac{-2xyf'}{f^2}\right ) + \frac{1}{y}\left(\frac{f-yf'2y}{f^2}\right)=\frac{1}{fy}.$$
$$\frac{1}{x}\left (\frac{-2xyf'}{f^2}\right ) + \frac{1}{y}\left(\frac{-yf'2y}{f^2}\right)=0$$
$$-\left (\frac{2yf'}{f^2}\right ) + \left(\frac{-2yf'}{f^2}\right)=0$$
Are you sure it's $f(x^2+y^2)$ and not $f(x^2-y^2)$ ???
