How to find all involutions on $\mathbb{C}(x,y)$, or at least all involutions $\delta$ on $\mathbb{C}(x,y)$ such that $\delta(x)=x$?

Remarks:

(1) An involution on $\mathbb{C}[x,y]$ is either conjugate to $\beta: (x,y) \mapsto (x,-y)$ or to $\epsilon: (x,y) \mapsto (-x,-y)$ (since the group of automorphisms of $k[x,y]$ is free amalgamated, and by a result of Serre about trees).

(2) In $\mathbb{C}[x,y]$ if we wish to find all involutions $\delta$ with $\delta(x)=x$, just write $\delta(y)=q$. There are two cases: (a) The Jacobian of $\delta$ is $1$, so $q_y=1$, and then $q=y+H(x)$, for some $H(t) \in \mathbb{C}[t]$. Since $\delta$ is of order $2$, by $\delta^2=1$ (=order $\leq 2$), we get $h(x)=0$, so $\delta=1$, which is not an involution. (b) The Jacobian of $\delta$ is $-1$, so $q_y=-1$, and then $q=-y+H(x)$, for some $H(t) \in \mathbb{C}[t]$. Since $\delta$ is of order $2$, by $\delta^2=1$ (=order $\leq 2$), we get that there is no restriction on $h(x)$, so $\delta: (x,y) \mapsto (x,-y+H(x))$. (Another way to obtain this is by using Theorem 1).

(3) Of course, in $\mathbb{C}(x,y)$ there exist more involutions, for example, $(x,y) \mapsto (x,\frac{1}{y})$. Is it true that the $\delta$'s I am looking for are those of (2)(b), $(x,y) \mapsto (x,\frac{1}{y})$ and $(x,y) \mapsto (x,-\frac{1}{y})$?

(4) I have asked this question, in which commenters suggested the following references: Serge Cantat's notes and Julie Deserti's list. (Perhaps I can find an answer to my recent question in those references).

Thank you very much.