An differential equation$(x^{2}＋y^{2}＋3)\frac{dy}{dx}＝2x(2y-\frac{x^{2}}{y})$ How to solve this ODE $$(x^{2}＋y^{2}＋3)\frac{dy}{dx}＝2x(2y-\frac{x^{2}}{y})$$
I tried to find its integral factor, but failed.
Many thanks for your help.

$(y^2+x^2+3)(y^2)'=4x(2y^2-x^2)$
$(y^2+x^2+3)(y^2+x^2+3)'=4x(2y^2-x^2)+2x(y^2+x^2+3)$
$(y^2+x^2+3)(y^2+x^2+3)'=10x(y^2+x^2+3)-4x(3x^2+6)$
With $\enspace z:=y^2+x^2+3\enspace$ we get $\enspace zz'=10xz-12x^3-24x$ . 
I don't know how to solve this, but Wolfram solves it in a closed form with  
http://www.wolframalpha.com/input/?i=z(dz%2Fdx)%3D10xz-12x%5E3-24x .
