hint only
$p(x,y)\in\mathbb{R}[x,y]$ and $\forall x,y\not=0:$ $p(x,y)=p(xy,\frac{1}{y})$ $(*)$. prove $\exists$ $s(x,y),r(x,y)$ that apply to the Eq $(*)$ and $\exists$ $q(x,y)$ so that
$$p(x,y)=q(s(x,y),r(x,y))$$
my attempt : $$p(x,y)=\sum_{0\leq{m,n}\leq{N}}a_{mn}x^my^n=\sum_{0\leq{m,n}\leq{N}}a_{mn}(xy)^m\left(\frac{1}{y}\right)^n$$ multiply by $y^N$ $$\sum_{0\leq{m,n}\leq{N}}a_{mn}x^my^{n+N}=\sum_{0\leq{m,n}\leq{N}}a_{mn}x^my^{m+N-n}$$ suppose that $(n>m)$ then the max power of $y$ in the LHS is $(N)$ and in RHS is $(N+(n-m))$ but $(n>m)$ therefor $\forall n>m:a_{mn}=0$
now we can conclude that $$p(x,y)=\sum_{0\leq{m,n}\leq{N}}a_{mn}(xy)^nx^{(n-m)}=Q(xy,x)$$ $$p(x,y)=p(xy,\frac{1}{y})=Q(x,xy)$$ $$Q(xy,x)=Q(x,xy)$$ therefor $Q$ is a symmetric polynomial