ODE $\frac{(f')^{2} f''}{f^3} = \lambda$ I am trying to solve the following PDE by separation of variables using $u(x, y) = v(x) + w(y)$
$\left(\frac{\partial u}{\partial x}\right)^2 \frac{\partial^2 u}{\partial {x^2}} + 
\frac{\partial u}{\partial x} \frac{\partial u}{\partial y}
\frac{\partial^2 u}{\partial x \partial y} + 
\left(\frac{\partial u}{\partial y}\right)^2 \frac{\partial^2 u}{\partial {y^2}} = 0
$
Inserting yields:
$ (v')^2 v'' w^3 + 0 + (w')^2 w'' v^3 = 0$
By dividing by $w^3$ and $v^3$ it follows that (in points $(x,y)$ where $v(x) \neq 0$ and $w(y) \neq 0$)
$\frac{(v')^2v''}{v^3}= \lambda$ for some constant $\lambda$
Which is equivalent to $(v')^3 v'' = \lambda v^3 v' \Leftrightarrow ((v')^4)' = \lambda (v^4)'$
Integrating the last equation gives us
$(v')^4 = \lambda v^4 + C$ for some constant $C$
In case $C = 0$ (i.e. when $\lim_{\lvert x \rvert \rightarrow\infty} v(x), v'(x) = 0$) then the solution is 
$v(x) = \alpha \exp(\beta \; x)$ where $\alpha, \beta \in \mathbb{R}$ with $\beta^4=\lambda$
I have difficulties to solve this differential equation when $C$ is not $0$.
Maybe the differential equation can be rewritten/simplified in another way, i don't see?

A last note: If $\lambda = 0$ it follows that $v'' = 0$ thus $v$ is an arbitray affine linear function 
$v(x) = mx + x_0$ 
 A: $$\left(\frac{dv}{d\omega}\right)^4=\lambda v^4+C$$
$$\frac{dv}{d\omega}= (\lambda v^4+C)^{1/4}$$
$V=-\frac{\lambda}{C}v^4 \quad\to\quad \frac{dV}{d\omega} = -\frac{4\lambda}{C}v^3 \frac{dv}{d\omega} =4\left(-\frac{\lambda}{C} \right)^{1/4}V^{3/4}\frac{dv}{d\omega} = 4\left(-\frac{\lambda}{C} \right)^{1/4}V^{3/4}C^{1/4}(1-V)^{1/4}$
$$\frac{dV}{d\omega} =4(-\lambda)^{1/4}V^{3/4}(1-V)^{1/4}$$
$$\omega=\frac{1}{4(-\lambda)^{-1/4}} \int V^{-3/4}(1-V)^{-1/4}dV$$
$$\omega= \frac{1}{4(-\lambda)^{-1/4}}\text{B}\left(V\:;\:-\frac34\:,\:-\frac14\right)+c$$
B$(x;a,b)$ is the Incomplete Beta function : https://en.wikipedia.org/wiki/Beta_function#Incomplete_beta_function
The result, expressed on implicit form, is :
$$\omega(v)= \frac{1}{4(-\lambda)^{1/4}}\text{B}\left(-\frac{\lambda}{C}v^4\:;\:-\frac34\:,\:-\frac14\right)+c$$
The explicit form $v(\omega)$ requiers the inverse function.
There is a relationship with the Gaussian hypergeometric function :
$$\omega(v)=v\:C^{-1/4} \:\:_2F_1\left(\frac14\:,\:\frac14\:;\:\frac54\:;\: -\frac{\lambda}{C}v^4\right)+c$$
https://en.wikipedia.org/wiki/Hypergeometric_function
A: This is NOT an answer to the question itself which was the solving the ODE $\quad\frac{(f')^{2} f''}{f^3} = \lambda.$
I already gave an answer to the question itself. Now, I would like to make a point about the PDE (which is not the ODE, of course):
$$\left(\frac{\partial u}{\partial x}\right)^2 \frac{\partial^2 u}{\partial {x^2}} + 
\frac{\partial u}{\partial x} \frac{\partial u}{\partial y}
\frac{\partial^2 u}{\partial x \partial y} + 
\left(\frac{\partial u}{\partial y}\right)^2 \frac{\partial^2 u}{\partial {y^2}} = 0
$$
The OP tried to solve the PDE by separation of variables, using $\quad u(x, y) = v(x) + w(y)\quad$ and transformed it into : $\quad (v')^2 v'' w^3 + 0 + (w')^2 w'' v^3 = 0\quad$ which is false.
In fact, the correct transformation is : 
$u_x=v'\qquad;\qquad u_y=w'\qquad;\qquad u_{xx}=v''\qquad;\qquad u_{yy}=w''\qquad;\qquad u_{xy}=0.$
$ (v')^2 v'' + 0 + (w')^2 w'' = 0$
$$(v')^2 v'' = -(w')^2 w''=\lambda$$
$$\begin{cases} 
(v')^2 v'' = \lambda \quad\to\quad v(x)=\frac{3^{4/3}}{4\lambda}(\lambda x+c_1)^{4/3}\\
(w')^2 w''=-\lambda \quad\to\quad w(y)=-\frac{3^{4/3}}{4\lambda}(-\lambda y+c_2)^{4/3}
\end{cases}$$ 
