 Solution

The substitution gives $x^2-y^2=u^2\cos 2v$. For this function $f_{uu}=2\cos 2v$, $f_u/u=2\cos 2v$, and $f_{vv}/u^2=-4\cos 2v$, whose sum is zero. Thus it satisfies the polar form of Laplace's equation for $u>0$. The apparent coordinate singularity at $u=0$ does not make the original Cartesian polynomial singular; it is a smooth harmonic function through the origin.