Solve ODE $yy'y''=(y')^3+(y'')^2$ 
Solve ODE $yy'y''=(y')^3+(y'')^2$.

Attempt. $y=0$ is clearly a solution. If $y\neq 0$, then ODE becomes
$$y=\frac{(y')^2}{y''}+\frac{y''}{y'}.$$
Let $y'=u$, so by differentiating with respect to variable $x$
we derive an equation of the form $y'=f(y',y'',y''')$, i.e.
$u=f(u,u',u'')$ (non linear, i guess). Am I on the right path? Is there an algorithm for solving this category (if so) of equations?
Thanks in advance for the help.
 A: Sometimes the re-parametrization by $y$ and resulting reduction of order helps see useful patterns in the equation. Thus assume $$y'=u(y)\implies y''(x)=(u(y(x))'=u'(y(x))y'(x)=u'(y)u(y)$$ for some non-constant monotonous segment of a solution to find
$$
yu^2u'=u^3+u^2u'^2\iff u^2(yu'-u'^2-u)=0
$$
So for $u\ne 0$ you get a Clairaut equation for $u$
$$
u(y)=yu'(y)-u'^2.
$$
This is a standard example, the solutions for $u'=C$ are lines $u(y)=Cy-C^2$ so that the resulting equation
$$
y'=Cy-C^2
$$
is easy to solve, for the singular solution $u(y)=\frac{y^2}4$ the resulting
$$
y'=\frac{y^2}4
$$
is likewise easy to solve.
Note that Clairaut equations allow to change from the regular solution family to the singular solution/envelope and back, this results in likewise piecewise solutions to the original equation. Additionally changes to $u=0$, that is $y=const.$ are possible.

Having identified the "hidden" type as Clairaut, you can now also directly compute from the original equation in that the derivative of that equation has to factorize completely modulo the original equation (orig).
\begin{align}
&&y'^2y''+yy''^2+yy'y'''&=3y'^2y''+2y''y'''\\
&&(yy''-2y'^2)y''&=(2y''-yy')y''',\tag{deriv1}\\
\text{(orig)}\implies&&(yy'-2y'')y''&=y'^3-y''^2\\
&&(yy''-2y'^2)y'&=y''^2-y'^3\\
&&&=-(yy'-2y'')y''\\
\text{(deriv1)}\implies&& 0&=(2y''-yy')(y'y'''+y''^2)
\end{align}
where each factor leads to the previously found equations.
