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][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][2], 
in which commenters suggested the following references: [Serge Cantat's notes][3] and 
[Julie Deserti's list][4]. (Perhaps I can find an answer to my recent question in those references).


Thank you very much.

  [1]: http://www.ams.org/journals/proc/1995-123-10/S0002-9939-1995-1257100-4/S0002-9939-1995-1257100-4.pdf
  [2]: https://mathoverflow.net/questions/253481/finding-all-automorphisms-of-mathbbcx-y
  [3]: https://perso.univ-rennes1.fr/serge.cantat/Articles/Survey-Cremona-SLC.pdf
  [4]: https://webusers.imj-prg.fr/~julie.deserti/cremona.html