Consider two functions $g_1(x_1,x_2)$, $g_2(x_1,x_2)$. The following holds,
\begin{align} \dfrac{\partial g_1}{\partial x_1}g_1^2 + \dfrac{\partial g_2}{\partial x_2}g_2^2 +g_1 g_2 \left(\dfrac{\partial g_1}{\partial x_2}+\dfrac{\partial g_2}{\partial x_1}\right) =0 \end{align} if and only if \begin{align} \dfrac{\partial g_1}{\partial x_1}=0,~\dfrac{\partial g_2}{\partial x_2}=0,~\left(\dfrac{\partial g_1}{\partial x_2}+\dfrac{\partial g_2}{\partial x_1}\right)=0. \end{align} Is there a way to prove or disprove this??
PS: this is equivalent to saying $$g^\top(x_1,x_2) \nabla_{(x_1,x_2)}g(x_1,x_2) g(x_1,x_2)=0 \iff \nabla_{(x_1,x_2)}g(x_1,x_2)+\nabla_{(x_1,x_2)}^\top g(x_1,x_2) = 0,$$ where $g(x_1,x_2)=(g_1(x_1,x_2),g_2(x_1,x_2))$.
I have tried many different ways, but could not prove/disprove this. However, there is a lot of geometry in the pde.
Why do you think these two conditions should be equivalent? Without additional hypotheses, there might be many solutions to the first equation that don't satisfy the second set of equations. For example, in the first quadrant, take $$ g_1(x_1,x_2) = (x_1)^{1/3}, \qquad g_2(x_1,x_2) = -(x_2)^{1/3}. $$