Reduce to first order and solve $yy'' = 3y'^2$ Reduce to first order and solve $yy'' = 3y'^2$
Dividing both sides by $y$ and re-arranging  $y'' - 3y' = 0$. This is clearly an homogenous equation, a solution might be $e^{3x}$to test this:
$y' = 3e^{3x}, y'' = 9e^{3x} \implies 9e^x-9e^x = 0$
Substituting $y = ue^{3x}, y' = u'e^{3x}+3ue^{3x}, y'' = u''e^{3x}+3u'e^{3x}+3u'e^{3x}+9ue^{3x}$
Plugging these in $$(u''e^{3x}+3u'e^{3x}+3u'e^{3x}+9ue^{3x})-3\cdot (u'e^{3x}+3ue^{3x})=0$$
$$\implies u''e^{3x}+3u'e^{3x}=0$$
Substituting $v = u'$
$$\implies v'e^{3x}+3ve^{3x}=0$$
$$=\int\frac{dv}{v} = 3\int dx$$
However the solution to this exercise is $(c_1x+c_2)^{-\frac{1}{2}}$ , how did they get this?

Hint...you can't cancel $y$ like you have. Instead put $y''=y'\frac{dy'}{dy}$ and the result will follow.
