Third order nonlinear ODE

I found a solution to the ODE, but it does not fit all initial conditions.

Ignoring initial conditions, a solution to this ODE is $$y=\frac{-6c}{cx+d}. $$

How I found this solution: We have $y'''=yy''-(y')^2$. By induction, we can prove $$y^{(2n)}=\sum_{r=0}^{n}b_r^{2n}y^{(r)}y^{(2n-1-r)}$$ and $$y^{(2n+1)}=\sum_{r=0}^{n}b_r^{2n+1}y^{(r)}y^{(2n-r)}.$$ Now let $n=1$, (I know that it is not true, but it helps to find a solution.) we have $$y''=b_0^{2}yy'+b_1^{2}yy'=(b_0^{2}+b_1^{2})yy'=\frac{b_0^{2}+b_1^{2}}{2}(y^2)'$$ which has the solution of the form $$y=\frac{a}{cx+d}. $$ Plugging this $y$ in the ODE, shows that $a=-6c$.

THIS NOTE MAY HELP

We want to solve $$ y'''+(y')^2-yy''=0\tag 1 $$ Instead of (1) I will solve $$ y'''-(y')^2-yy''=0\tag 2 $$ We have $$ y'''-(y')^2-yy''=0\Leftrightarrow y'''-(yy')'=0\Leftrightarrow y''-yy'=-C_1\Leftrightarrow $$ $$ (y'-\frac{y^2}{2})'=(-C_1x)'\Leftrightarrow y'-y^2/2=-C_1x-C_2 $$ If we set $y=-2u'/u$ we arrive to $$ u''=\frac{1}{2}(C_1x+C_2)u $$ The last equation is solvable with Airy $\textrm{Ai}(x)$,$\textrm{Bi}(x)$ functions see Wikipedia. $$ y(x)=-2^{2/3}C_1^{1/3}\frac{\textrm{Bi}'\left(\frac{C_1x+C_2}{2^{1/3}C_1^{2/3}}\right)+\textrm{Ai}'\left(\frac{C_1x+C_2}{2^{1/3}C_1^{2/3}}\right)C_3}{\textrm{Bi}\left(\frac{C_1x+C_2}{2^{1/3}C_1^{2/3}}\right)+\textrm{Ai}\left(\frac{C_1x+C_2}{2^{1/3}C_1^{2/3}}\right)C_3} $$ For the conditions $y(0)=0$, $y'(0)=-1$, $y'(\infty)=0$, we easily get $C_1=1/2$,$C_2=1$,$C_3=-\textrm{Bi}'(2^{1/3})/\textrm{Ai}'(2^{1/3})$.