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!

Hint
$$2y'= {\frac{xy}{x^2+1}} - {\frac{2x}{y}}$$
You can also multiply by y the Bernouilli 's equation
$$2y'y= {\frac{xy^2}{x^2+1}} - 2x$$
Observe that $(y^2)'=2y'y$ substitute $z=y^2$
$$z'= {\frac{xz}{x^2+1}} - 2x$$
Now it's a linear first ode
But you can do it your way
$${\frac{2y'}{y}} ={\frac{x}{x^2+1}} - {\frac{2x}{y^2}}$$
$${\frac{2y'}{y^2}} ={\frac{x}{y(x^2+1)}} - {\frac{2x}{y^3}}$$
Substitute $z=1/y$
$$-2z' =\frac{zx}{(x^2+1)} - 2xz^3$$
