$\displaystyle (2x^2ylny-x)y'=y$
$\displaystyle (2x^2ylny-x)dy=ydx$
then i divide both sides by dy
$\displaystyle (2x^2ylny-x)=yx'$
then i divide both sides by y
$\displaystyle (2x^2lny-\frac{x}{y})=x'$
$\displaystyle x'+\frac{x}{y}=2x^2lny$
so i have here a bernuly foruma
i divide both sides by $\displaystyle x^2$
$\displaystyle \frac{x'}{x^2}+\frac{1}{xy}=2lny$
$\displaystyle z=x^{-1}$
$\displaystyle z'=-1x^{-2}x'$
$\displaystyle -z'+\frac{z}{y}=2ln y$
z is defined to be a function of x
so $\displaystyle z'=\frac{dz}{dx}$
why the book interprets $\displaystyle z'=\frac{dz}{dy}$
??