I think this question should be easy and shouldn't require me to solve the entire PDE for a general solution. Basically, how would you see immediately that the general solution of:
$$y^2 u_{xx}-2xy u_{xy}+x^2u_{yy}=y^2/x\ u_x+x^2/y \ u_x$$
Can be written as $u(x,y)=(x^2-y^2) f(x^2+y^2)+g(x^2+y^2)$ given that f and g are both arbitrary functions?
I'm not sure it's supposed to be that easy. The thing on the left is recognizable as the result of plugging the vector $(y,-x)$ into the Hessian of $u$. This looks like the second-order tangential derivative of $u$. One has to be careful not to mistake it for $u_{\theta\theta}$ (with second derivatives, the curvature of polar coordinates gets in the way). But at least we see that polar coordinates may help here.
So, use the standard conversion formulas: $$u_\theta = -yu_x+xu_y$$ $$u_{\theta\theta} = y^2 u_{xx} - 2xy u_{xy} + x^2 u_{yy} - xu_x-yu_y$$ The equation takes the form $$ u_{\theta\theta} = \frac{y^2}{x} u_x + \frac{x^2}{y} u_y - xu_x-yu_y $$ which simplifies to $$ u_{\theta\theta} = - u_\theta \cot 2\theta $$ This ODE can be solved, producing $A+B\cos 2\theta$. The constants $A,B$ can depend on $r$ in any way.