Let f be an arbitrary, twice differentiable function for which $f''\neq0$ Let f be an arbitrary, twice differentiable function for which $f''\neq0$ . The function $u(x,y)=f(x^2+axy+y^2)$ satifsfies equation $U_{xx}-U_{yy}=0$ then the constant a is. 
$f_{xx}(x^2+axy+y^2)(2x+ay)^2+2f_x(x^2+axy+y^2)-f_{yy}(x^2+axy+y^2)(2y+xa)^2-2f_y(x^2+axy+y^2)=0$

By the chain rule, $$U_{xx} = 2f'(x^2+axy+y^2) + (2x+ay)^2f''(x^2+axy+y^2)$$ and $$U_{yy} = 2f'(x^2+axy+y^2) + (2y+ax)^2f''(x^2+axy+y^2).$$
Since $U_{xx}-U{yy}=0$, and $f''(x) \neq 0, \forall x \in D_f$, we have $$((2x+ay)^2-(2y+ax)^2)f''(x^2+axy+y^2) = 0 \Rightarrow 4x^2+4axy+a^2y^2 - 4y^2 -4axy - a^2x^2 = 0.$$
In particular, taking $x=1$ and $y=0$, we have $$4-a^2= 0$$ therefore the answer, if it exists, must be $2$ or $-2$ but checking on the original equation, we have that both $a = 2$ and $a = -2$ are solutions.
