Let $k$ be a field of characteristic zero. Let $\beta: k[x,y] \to k[x,y]$ be the involution (a $k$-algebra automorphism of order two) defined by $(x,y) \mapsto (x,-y)$.
The set of symmetric elements with respect to $\beta$, denote it by $S_\beta$, is of the form $a_{2n+2}y^{2n+2}+a_{2n}y^{2n}+\cdots+a_2y^2+a_0$, where $a_{2n+2},a_{2n},\ldots,a_2,a_0 \in k[x]$.
The set of skew-symmetric elements with respect to $\beta$, denote it by $K_\beta$, is of the form $a_{2n+1}y^{2n+1}+a_{2n-1}y^{2n-1}+\cdots+a_3y^3+a_1y$, where $a_{2n+1},a_{2n-1},\ldots,a_3,a_1 \in k[x]$.
Clearly, $S_\beta=k[x,y^2]$. ($K_\beta$ is not a subalgebra of $k[x,y]$, since, for example, the product of two skew-symmetric elements is a symmetric element).
Assume that $s_1,s_2 \in S_{\beta}$ are such that $k(s_1,s_2)=k(x,y^2)$.
In this situation:
Question 1: Is it possible to find the 'exact' form of such $s_1,s_2$?
(An attempt to answer Question 1: We can write $x=G(s_1,s_2)$ and $y=H(s_1,s_2)$, for some $G,H \in k(U,V)$. Then perhaps (if not dividing by zero) we can substitute $(x,y)$ by $(x,0)$ and $(0,y)$. Also, we can differentiate by $x$ and by $y$).
Example: $s_1=y^2+x^2+x$ and $s_2=y^2+x^2$; in this case actually, $k[s_1,s_2]=k[x,y^2]$.
This example leads to the following question:
Question 2: Is it true that $k(s_1,s_2)=k(x,y^2)$ implies that $k[s_1,s_1]=k[x,y^2]$?
Notice that taking $s_1=x^2,s_2=x^3$ is not a counterexample to Question 2, because $k(s_1,s_2)=k(x^2,x^3)=k(x) \subsetneq k(x,y^2)$.
Any hints and comments are welcome!