hint:$s(x,y)=x^2y$,$r(x,y)=x+xy$.similar problem

"In the name of God" Firstly, let good $p(x,y)$ takes the form: $p(x,y) = \sum_{m,n}^{ } \left ( a_{m,n} .x^{m} .y^{n} \right )$ where $a_{m,n}$ is a real number and $m,n$ are integers such that: $0 \leq m,n < N $. Main assertion implies that $p(x,y) = p\left(xy,\frac{1}{y}\right)$ ;therefore, $\sum_{m,n}^{ } \left ( a_{m,n} .x^{m} .y^{n} \right )$ =$\sum_{m,n}^{ } \left ( a_{m,n} .x^{m}y^{m} .y^{-n} \right )$. By multiplying both sides by $y^N $we obtain: $\sum_{m,n}^{ } \left ( a_{m,n} .x^{m} .y^{n+N} \right )$ =$\sum_{m,n}^{ } \left ( a_{m,n} .x^{m} .y^{m+N-n} \right )$.Claim: $n \leq m$ . Proof, for the sake of contradiction assume the opposite. Thus, the degree of $y$ in $R.H.S.$ is atleast $N$. Measuring the degree of $y$ in $L.H.S.$ would give us the contradiction. Hence $n \leq m$ and $\sum_{m,n}^{ } \left ( a_{m,n} .x^{m} .y^{n} \right )$ =$\sum_{m,n}^{ } \left ( a_{m,n} .x^{m}y^{m} .y^{m-n} \right )$. Now we define a $q(x,y)\in \mathbb{R}\left[x,y\right]$ such that: $p(x,y) = q(xy,x) $ and rewriting the main assertion with $q(x,y)$ implies that(let $xy=z$): $q(x,z)=q(z,x)$ so $q(x,y)$ is symmitric. Lemma: For each polynomial $q(x,y)$ satisfying$q(x,y)=q(y,x)$ therefore exist a polynomial $f(x,y)\in \mathbb{R}\left[x,y\right]$ such that $q(x,y)=f(x+y,xy) $proof: by easy induction. Using this lemma implies that $p(x,y)=q(xy,x)=f(xy+x,x^2y)$. So it's easy to check that $s(x,y)=x^2y$ and $r(x,y)=x+xy$ satisfies the assertion. So we're done!