Solution of the first order differential equation $y'=\frac{x(x^2+y^2)^2}{4y}$. I m stuck in finding the solution of the following differential equation 
$$y'=\frac{x(x^2+y^2)^2}{4y}.$$
Please give me some hint.

$\frac{4yy'}{(x^2+y^2)^2}=x$ 
then define $x^2+y^2=h^2 \Rightarrow 2hh'=2yy'+2x \Rightarrow y'=\frac{hh'-x}{y}$ 
then by substituting we get: $\frac{4(hh'-x)}{h^4}=x \Rightarrow4(hh'-x)=xh^4 \Rightarrow \frac{4hh'}{4+h^4}=x$
now define $g=h^2$ and solve the problem: $\Rightarrow g'=2hh'$
$$\frac{2g}{4+g^2}=x \Rightarrow \tan^{-1}(\frac{g}{2})=\frac{x^2}{2}+C \Rightarrow g=2\tan(\frac{x^2}{2}+C)$$
then we get: $$h=\pm \sqrt{2\tan(\frac{x^2}{2}+C)} \Rightarrow y=\pm \sqrt{ 2\tan(\frac{x^2}{2}+C)-x^2}$$
