Method 1:
Let y'=p(y), then y''=p'*dy/dx=p'p, so the original formula is: yp' p+p 2 = 0, that is, yp' =-p.
After separating the variables: dp/p=-dy/y, the two-side integration results in: ln|p|=-ln|y|+ln|C 1|, that is, p = c1/y.
Then dy/dx=C 1/y, and then the variables are separated: ydy=C 1dx.
The integrals on both sides are1/2y 2 = c1x+C2, that is, y 2 = c3x+C4.
The initial conditions are your own.
Method 2: y'' y+(y') 2 = 0 Deduce (yy')'=0, then yy'=C 1, then 1/2 (y 2)' = c 1.
Then (y 2)' = C2, then y 2 = C2X+C3.