$$y''=y'y \implies y''=\frac 12 (y^2)' $$
After integration
$$\implies y'=\frac 12 y^2+K$$
$$y'=\frac 12(y^2+2K)$$
For $ K=0 \implies -\frac 1y=\frac x2 +a$
$$\implies y(x)=-\frac 2 {x+c}$$
For K negative
$$ \int \frac {dy}{y^2-c^2}=\frac x 2+b$$
$$\ln (\frac {y-c}{y+c})=cx+a$$
$$y(x)=-c \frac {ae^{cx}+1}{ae^{cx}-1}$$
For K positive, substitute $2K=c^2$
$$ \int \frac {dy}{y^2+c^2}=\frac 12\int dx=\frac x 2+b$$
Substitute $z=y/c \implies dz=dy/c$
$$ \int \frac {dz}{z^2+1}=c(\frac x 2+b)$$
$$ \arctan (z)=c(\frac x 2+b)$$
$$ y=c\tan (c(\frac x 2+b))$$
Substitute $c/2=k$ and $bc=a$
$$ y=2k\tan (kx+a)$$
$$y''=y'y$$
Substitute
$$y'= p \implies y''=pp'$$
$$p'=y \implies p=\frac 12 y^2 +K$$
$$y'=\frac 12 y^2+K$$
This equation is separable