Non-linear differential equation that allows reducing 
I have been trying to solve a bunch of non-linear ordinary equations, but each of them presents a problem while separating variables.   After recommended substitution y' = p I have

What would my further steps look like? Thank you in advance
 A: A key observation is the absence of $t$. Using the chain rule, the substitution $p = \dot{y} = \frac{dy}{dt}$ makes the second derivative of $y$ $$\ddot{y} = \frac{dp}{dt} = \frac{dp}{dy}\frac{dy}{dt} =  \frac{dp}{dy}p$$ and after substituting in the ODE (with independent variable $y$) only the first derivative of $p$ will appear. In particular, in the original ODE $y\dot{y}\ddot{y} = \dot{y}^3 + \ddot{y}^2$, denote $\frac{dp}{dy}$ by $p'$ we obtain $$y p^2 p' = p^3 + p'^2p^2$$
Assuming that $p \neq 0$, then
$$yp'  = p + p'^2 \tag 1$$
This is the reduced nonlinear first order ODE. Notice if we assume a solution $p = ay - b$ then it indeed solves the reduced ODE, substituting in (1)  $$ay = ay -b + a^2 \quad \implies \quad  a = \sqrt{b}$$
Back to the definition of $p = \dot{y}$, we need to solve (in t) the non-homogeneous linear ODE $$\dot{y} = \sqrt{b}y - b$$
The solution is routine, and is $$y(t) = y_0 e^{\sqrt{b}t} + \sqrt{b} \implies \dot{y}(t) = y_0 \sqrt{b}e^{\sqrt{b}t} \implies \ddot{y}(y) = y_0 b e^{\sqrt{b}t}$$
Finally, indeed
$$y\dot{y}\ddot{y} = y_0^3 \sqrt{b}b e^{3\sqrt{b}{t}} + y_0^2b^2 e^{2\sqrt{b}{t}} = \dot{y}^3 + \ddot{y}^2$$
