Is this (1) y"=-½y2 or (2) y"=-1/(2y2)? Both possibilities will be considered.
Let z=dy/dx=y', y"=z'; (dz/dy)(dy/dx)=dz/dx; y"=z'=dz/dx=z(dz/dy).
(1) z(dz/dy)=-½y2.
∫zdz=-½∫y2dy, z2/2=-⅙y3+a where a is a constant.
z2=-⅓y3+a, dy/dx=√(a-⅓y3).
y'(0)=-1, so -1=√(a-⅓(y(0))3)=√(a-⅓), because y(0)=1, a-⅓=1, a=4/3.
∫dy/√(4/3-⅓y3)=∫dx. This may not be conventionally integrable.
(2) z(dz/dy)=-1/(2y2).
z2/2=-½∫dy/y2, z2=-∫y-2dy,
z2=1/y+a, y'=z=√(1/y+a) or y'=-√(1/y+a). y(0)=1, y'(0)=-1⇒a=0. Only the negative square root applies.
dy/dx=-y-½, because a=0.
∫y½dy=-∫dx, ⅔y3/2=-x+b, so b=⅔, because y(0)=1, and ⅔y3/2=⅔-x, or y3/2=1-3x/2.
CHECK
y'√y=-1, y"√y+y'(y'/(2√y)=0, y"√y+(-1/√y)2(1/2√y)=0, y"+1/(2y2)=0, y"=-1/(2y2). Checks out.
It would appear that the solution to y"=-1/(2y)2 is y3/2=1-3x/2, or maybe y=∛(9x2/4-3x+1). However, these two equations are not exactly equivalent, though similar, as can be clearly seen if the equations are graphed. This is due to the fact that only the square roots of positive numbers are valid; y=∛(9x2/4-3x+1) contains two symmetrical halves, while y3/2=1-3x/2 only has the left half of the same curve. For example, there is no real value for y when x>⅔. The unique solution is therefore y3/2=1-3x/2.