Perhaps the following question is not in the level of MO questions, but it has not received comments in MSE, so I ask it here also:
Let $\beta: \mathbb{C}[x,y] \to \mathbb{C}[x,y]$ be the involution on $\mathbb{C}[x,y]$ defined by $(x,y) \mapsto (x,-y)$.
Let $s_1,s_2,s_3 \in \mathbb{C}[x,y]$ be three symmetric elements with respect to $\beta$. It is not difficult to see that a symmetric element w.r.t. $\beta$ is of the following form: $a_{2n}y^{2n}+a_{2n-2}y^{2n-2}+\cdots+a_2y^2+a_0$, where $a_{2j} \in \mathbb{C}[x]$.
Assume that the following two conditions are satisfied:
(1) Each two of $\{s_1,s_2,s_3\}$ are algebraically independent over $\mathbb{C}$. Notice that the three $s_1,s_2,s_3$ are algebraically dependent over $\mathbb{C}$, since the transcendence degree of $\mathbb{C}[x,y]$ over $\mathbb{C}$ is two.
(2) $\mathbb{C}(s_1,s_2,s_3,y)=\mathbb{C}(x,y)$; this notation means the fields of fractions of $\mathbb{C}[s_1,s_2,s_3,y]$ and $\mathbb{C}[x,y]$, respectively.
Example: $s_1=x^2+x^5+A(y), s_2=x^5y^2+B(y), s_3=x^3y^2+C(y)$, where $A(y),B(y),C(y) \in \mathbb{C}[y^2]$.
Question 1: Is it possible to find a 'specific' form of at least one of $\{s_1,s_2,s_3\}$?
A plausible answer may be: One of $\{s_1,s_2,s_3\}$ is of the form
$\lambda x^ny^{2m}+D(y)$ for some $D(y) \in \mathbb{C}[y^2]$, $\lambda \in \mathbb{C}^{\times}$, $n \geq 1$, $m \geq 0$; is it possible to find a counterexample to my plausible answer?
Perhaps it is better to first consider two (easier) questions replacing conditions (1) and (2) by:
(1') $\{s_1,s_2\}$ are algebraically independent over $\mathbb{C}$ + (2') $\mathbb{C}(s_1,s_2,y)=\mathbb{C}(x,y)$; call this Question 1'.
(1'') $s_1 \neq 0$ + (2'') $\mathbb{C}(s_1,y)=\mathbb{C}(x,y)$; call this Question 1''. I guess that the answer to question 1'' is: $s_1=\lambda xE(y) + F(y)$, where $\lambda \in \mathbb{C}^{\times}$ and $E(y),F(y) \in \mathbb{C}[y^2]$.
Remarks:
(i) In the above example we already have $\mathbb{C}(s_2,s_3,y)=\mathbb{C}(x,y)$ and $\mathbb{C}(s_1,s_2,y)=\mathbb{C}(x,y)$.
(ii) We can write $x=\frac{u(s_1,s_2,s_3,y)}{v(s_1,s_2,s_3,y)}$ for some $u,v \in \mathbb{C}[X,Y,Z,W]$. Then, if I am not wrong, taking $y=0$ (if possible?) we obtain that $x=\frac{u(s_1(x,0),s_2(x,0),s_3(x,0))}{v(s_1(x,0),s_2(x,0),s_3(x,0))}$, hence $\mathbb{C}(s_1(x,0),s_2(x,0),s_3(x,0))=\mathbb{C}(x)$.
Question 2: Is there an example where all $s_1,s_2,s_3$ are necessary to obtain $\mathbb{C}(s_1,s_2,s_3,y)=\mathbb{C}(x,y)$? Namely, it is not possible to omit one of $\{s_1,s_2,s_3\}$ and still get $\mathbb{C}(x,y)$. I guess that the answer is positive.
Thank you very much!
