This is just a long comment, but note that a symmetric polynomial in 2 variables, might not even be expressible as a linear combination of squares of symmetric polynomials. For example, $x^2+xy+y^2$ is not a linear combination of $(x+y)^2$ (the only possible option).

However, if $f(x_1,\dotsc,x_n)$ is a shift-invariant polynomial, i.e. $f(x_1,\dotsc,x_n) = f(x_1+t,\dotsc,x_n+t)$ for all $t$ (the discriminant is a natural example of such a polynomial), then (with the obvious constraint that it is of homgeneous even degree) it is a linear combination of squares of shift-invariant symmetric polynomials.

(The proof boils down to finding an explicit basis the vector space consisting of squares.)

So, perhaps in the symmetric and shift-invariant case, it might be possible that the rational functions themselves can be taken to be symmetric and shift-invariant...

This is just a long comment, but note that a symmetric polynomial in 2 variables, might not even be expressible as a linear combination of squares of symmetric polynomials. For example, $x^2+xy+y^2$ is not a linear combination of $(x+y)^2$ (the only possible option).

However, if $f(x_1,\dotsc,x_n)$ is a shift-invariant polynomial, i.e., $f(x_1,\dotsc,x_n) = f(x_1+t,\dotsc,x_n+t)$ for all $t$ (the discriminant is a natural example of such a polynomial), then (with the obvious constraint that it is of homgeneous even degree) it is a linear combination of squares of shift-invariant symmetric polynomials.

(The proof boils down to finding an explicit basis the vector space consisting of squares.)

So, perhaps in the symmetric and shift-invariant case, it might be possible that the rational functions themselves can be taken to be symmetric and shift-invariant...