$\displaystyle (2x^2ylny-x)y'=y$

$\displaystyle (2x^2ylny-x)dy=ydx$

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 $\displaystyle z'=\frac{dz}{dy}$

??