Solution of $\frac{\partial f}{\partial x}\frac{\partial f}{\partial y} - 2 f \frac{\partial^2 f}{\partial x \, \partial y} = 0$ I've been trying to to solve the following PDE:
\begin{equation}
\frac{\partial f}{\partial x}\frac{\partial f}{\partial y} - 2 f \frac{\partial^2 f}{\partial x \, \partial y} = 0
\end{equation}
I encountered this as a part my research and the only condition $f$ has to satisfy is that $f \rightarrow 0$ as $|(x,y)| \rightarrow \infty$. I would like to know how to determine if it is possible to find a non-trivial $f$ that satisfies the above (analytic solution would be great but not necessary).
What I have tried: 
-Any $f$ of the form $f(x,y) = (A(x) + A(y))^2$ will satisfies the equation but not the condition that $f$ tends to zero. 
-I tried NDSolve in Mathematica but didn't really get any conclusion I think mainly because I don't really know how to use it properly. 
 A: Firstly, the PDE can be made homogenous by letting $f=\exp{g(x,y)}$, where $g(x,y)$ must diverge to $-\infty$ as $\|(x,y)\|\to\infty$ so as to satisfy the condition for $f$. Rewriting the PDE,
\begin{equation}
\frac{\partial g}{\partial x}\frac{\partial g}{\partial y}+2\frac{\partial^2 g}{\partial x\, \partial y}=0.
\end{equation}
Secondly, since the PDE is symmetric with respect to the interchange between $x$ and $y$, its solution $f(x,y)$ must also be symmetric, and so must $g(x,y)$. This symmetry implies that the partial derivatives of $g(x,y)$ with respect to either $x$ and $y$ are equal,
\begin{equation}
\frac{\partial g(x,y)}{\partial x}=\frac{\partial g(y,x)}{\partial y}=\frac{\partial g(x,y)}{\partial y}.
\end{equation}
Since the partial derivatives are the same, $g(x,y)$ can be let equal to a single variable function $h(r)$, where $\frac{\partial r}{\partial x}=\frac{\partial r}{\partial y}=1$, i.e. $r=x+y$. Thus, the solution to the PDE lies in the solution to
\begin{equation}
\left(\frac{\partial h}{\partial r}\right)^2 + 2 \frac{\partial^2 h}{\partial r^2}=0.
\end{equation}
Unfortunately, I don't think there is a solution to this equation that satisfies $h(r)\to -\infty$ as $r\to\infty$. If this condition is dropped, then the solution is of the form
\begin{equation}
h(r)=2\log({|r+C_1|})+C_2,
\end{equation}
which upon back substitution gives a solution similar to what the OP has figured,
\begin{equation}
f(x,y)=(A(x)+A(y))^2.
\end{equation}
I hope this inconclusive working helps.
