Solve $y''=y^2$

Multiplying this equation by $2y'$, we get $$\left(y'^2\right)'=\frac23\left(y^3\right)',$$ which implies that $$y'^2=\frac23y^3+C_1.$$ Although the general solution of this can be written in terms of elliptic integrals (the equation is separable), there is a particularly simple solution corresponding to $C_1=0$: then $$y'=\pm \sqrt{\frac23}y^{\frac32}\qquad \Longrightarrow \qquad -2y^{-\frac12}=\pm \sqrt{\frac23} x+C_2.$$

A hint: Multiply both sides of the equation by $y'$.

A physicist's substitution always helps when the independent variable is missing and you have second derivatives. Let $t$ be the independent variable ($y'=dy/dt$). Then use a new variable $$y'=v$$ and substitute $$y''=v\,dv/dy$$ and integrate with respect to $y$.

In this case: $$v\,dv/dy=y^2$$ $$\int v\, dv=\int y^2 \, dy$$ $$v^2/2=y^3/3+C$$ Now put $v=dy/dt$ back in and integrate again to get $y(t)$.

In physics, you often have second derivatives over time (2nd Newton's law), so this is a routine bread-and-butter method for converting equations to first order and solving them routinely.