Solve the differential equation 2y'=yx/(x^2 + 1) - 2x/y I have to solve the equation:
$$ 2y'= {\frac{xy}{x^2+1}} - {\frac{2x}{y}} $$
I know the first step is to divide by y, which gives the following equation:
$$ {\frac{2y'}{y}} ={\frac{x}{x^2+1}} - {\frac{2x}{y^2}}$$
According to my notes I get that I should make a substitution:
 $$ z = {\frac{1}{y^2}} $$
and  the derivative of z:$$ z' = {\frac{y'}{y^3}}$$
But I don't know how to proceed after this... Any help is appreciated!

write your equation in the form
$$2\frac{dy(x)}{dx}y(x)-\frac{xy(x)^2}{x^2+1}=-2x$$
substituting
$$v(x)=y(x)^2$$ and $$\mu(x)=e^{\int\frac{-x}{\sqrt{x^2+1}}dx}=\frac{1}{\sqrt{x^2+1}}$$
we get
$$\frac{\frac{dv(x)}{dx}}{\sqrt{x^2+1}}-\frac{xv(x)}{\sqrt{x^2+1}^{3/2}}=\frac{2x}{\sqrt{x^2+1}}$$
and we get
$$\frac{\frac{d v(x)}{dx}}{\sqrt{x^2+1}}+\frac{d}{dx}\left(\frac{1}{\sqrt{x^2+1}}\right)v(x)=-\frac{2x}{\sqrt{x^2+1}}$$
and now $$\int\frac{d}{dx}\left(\frac{v(x)}{\sqrt{x^2+1}}\right)dx=\int-\frac{2x}{\sqrt{x^2+1}}dx$$
