Let $\beta: (x,y) \mapsto (x,-y)$, extend it to all of $\mathbb{C}[x,y]$ and get an involution (= an automorphism of order two) $\beta$ on $\mathbb{C}[x,y]$.
It is clear that every symmetric element with respect to $\beta$ is of the form: $h_{2m} y^{2m}+ h_{2m-2} y^{2m-2} +\cdots+ h_2 y^2+ h_0$, where $h_{2m},h_{2m-2},\ldots,h_2,h_0 \in \mathbb{C}[x]$.
The set of symmetric elements with respect to $\beta$, denote it by $S_{\beta}$, is generated by $x$ and $y^2$; in other words: $S_{\beta}=\mathbb{C}[x,y^2]$.
Assume that $f,g\in \mathbb{C}[x,y]$ are algebraically independent over $\mathbb{C}$; this is equivalent to saying that their Jacobian, $Jac(f,g):=f_xg_y-f_yg_x$, is non-zero.
Further assume that $f,g \in S_{\beta}$, namely, $f$ and $g$ are symmetric w.r.t. $\beta$.
Is it possible to determine when $\mathbb{C}(f,g) \subseteq \mathbb{C}(x,y^2)$ is Galois?
Remarks: (1) $\mathbb{C}(x,y^2) \subset \mathbb{C}(x,y)$ is Galois, since it is of order two. Its Galois group is $\{1,\beta\}$, where $\beta$ is the former $\beta$ extended to $\mathbb{C}(x,y)$.
(2) For example, $f=x$, $g=y^4$. $\mathbb{C}(x,y^4) \subset \mathbb{C}(x,y^2)$ is Galois, since it is of order two. More generally, if I am not wrong, $\mathbb{C}(x,y^{2n}) \subset \mathbb{C}(x,y^2)$ is Galois, since it is cyclic of order $n$.
I’m not sure that your $\beta$ is an essential part of your question. You’re pretty much just asking when a subfield $k$ of $K=\Bbb C(x,Y)$ that’s of transcendence degree two over $\Bbb C$ has $K\supset k$ Galois. (Here, $Y=y^2$.)
You might as well turn the question around and ask when a finite extension $K$ of $k=\Bbb C(f,g)$ is Galois. (Here, of course, $\,f$ and $g$ are any two algebraically independent things, may as well be taken to be two different indeterminates.)