The identity is impossible. In other words, there does not exist any $g(y)$ or $h(y)$ such that the equation $2x^2 + xg(y) + h(y) = 0 $ is true for all $x$ and $y$ on the domain of $u(x,y)$.

*Proof:*

Denote $v(x,y) = 2x^2 + xg(y) + h(y)$. Say we find some value $x=x_0$ and $y=y_0$ such that $v(x_0, y_0) = 0$. If $g(y_0)$ is positive or 0, let $x_1 > x_0$. Now closely examine the expression $v(x_1, y_0)$ and compare to $v(x_0, y_0)$. $2x_{1}^2 > 2x_{0}^2$ and $x_{1}g(y_0) > x_{0}g(y_0)$. Then $v(x_1, y_0) \ne v(x_0, y_0) = 0$.

If $g(y_0)$ is negative or 0, let $x_2 < x_0$. We would similarly find $v(x_2, y_0) \ne v(x_0, y_0) = 0$.

Thus, there is always some set of points $x$ and $y$ where $v(x,y) \ne 0$.

*How this applies to the problem (14)*:

The fact that the identity is never possible means that there does not exist a general solution for the constraints $u_{xx} = y^2$ and $u_{yy} = -x^2$.