Prove that $f(x,y)=g(x+y,xy)$. Let $f\in \mathbb R[x,y]$ such that $f(x,y)=f(y,x)$. Prove that $f(x,y)=g(x+y,xy)$.
My try.
$f\in \Bbb R[x,y]$ and $f(x,y)=f(y,x)$ .If I take for example $f(x,y)=a_{00}+a_{01}y+a_{11}xy+a_{10}x+a_{13}xy^3$ then only the coefficients which are symmetric w.r.t $x $ and $y$ will remain in the expression of $f$ ,all other terms will cancel out because $f(x,y)=f(y,x)$.
But How can I write down the equation of $f$ from here?
Please provide some hints.

Let $f\in \mathbb R[x,y]$ such that $f(x,y)=f(y,x)$. To prove that $f(x,y)=g(x+y,xy)$.
Let $$f(x,y)=a_{0,0}x^0y^0+a_{0,1}x^0y^1+a_{0,2}x^0y^2+a_{1,0}x^1y^0+a_{1,1}x^1y^1+a_{1,2}x^1y^2+a_{2,0}x^2y^0+a_{2,1}x^2y^1+a_{2,2}x^2y^2$$
We then have 
$$f(y,x)=a_{0,0}y^0x^0+a_{0,1}y^0x^1+a_{0,2}y^0x^2+a_{1,0}y^1x^0+a_{1,1}y^1x^1+a_{1,2}y^1x^2+a_{2,0}y^2x^0+a_{2,1}y^2x^1+a_{2,2}x^2y^2$$
As $f(x,y)=f(y,x)$ we have $a_{0,1}=a_{1,0}, a_{2,0}=a_{0,2},a_{1,2}=a_{2,1}$.. So, we have
$$f(x,y)=a_{0,0}x^0y^0+a_{0,1}x^0y^1+a_{0,2}x^0y^2+a_{0,1}x^1y^0+a_{1,1}x^1y^1+a_{1,2}x^1y^2+a_{0,2}x^2y^0+a_{1,2}x^2y^1+a_{2,2}x^2y^2=a_{0,0}x^0y^0+a_{0,1}(x^0y^1+x^1y^0)+a_{0,2}(x^0y^2+x^2y^0)+a_{1,1}x^1y^1+a_{1,2}(x^1y^2+x^2y^1)+a_{2,2}x^2y^2=a +bxy+c(xy)^2+d(x+y)+e((x+y)^2-2xy)+fxy(x+y)$$
