$\displaystyle y''y=(y')^{2}$

$\displaystyle y(0)=1 y'(0)=2$

$\displaystyle y'=z$

$\displaystyle z'y=z^{2}$

$\displaystyle \frac{{dz}}{dx}=\frac{dz}{dy}\frac{dy}{dx}$

$\displaystyle zz'=dz/dx$

i know i need to do x'(y)

but i dont know how to put it there

how i solve it?