Find general solution of $xx''=(x')^3$
$$\frac{x''}{x'^2}=\frac{x'}{x}$$ yields
$$-\frac1{x'}=\log x+c.$$
Then is suffices to integrate
$$(\log x+c)x'=-1.$$
$$\frac{x''}{x'^2}=\frac{x'}{x}$$ yields
$$-\frac1{x'}=\log x+c.$$
Then is suffices to integrate
$$(\log x+c)x'=-1.$$