 Solution

Choose a coprime presentationFor a polynomial $r$, write $r^{\ast }(X,Y)=r(Y,X)$. Symmetry of $f$ givesSince $\mathbb{C}[X,Y]$ is a unique factorization domain and $\gcd (p,q)=1$, one has $p\mid p^{\ast }$ and $q\mid q^{\ast }$. Exchanging the variables preserves total degree, sofor nonzero constants $\lambda ,\mu$. The displayed identity gives $\lambda =\mu$, and applying the exchange twice gives ${\lambda }^2=1$.

If $\lambda =-1$, then $p(X,X)=q(X,X)=0$, so $X-Y$ divides both polynomials, contradicting coprimality. Therefore $\lambda =1$, and both $p$ and $q$ are symmetric. Taking $g=p$ and $h=q$ proves the claim, which is the symmetric rational function in two variables result.