Let $f: (x,y) \mapsto (p,q)$ be a $\mathbb{C}$-algebra endomorphism of $\mathbb{C}(x,y)$ satisfying the following two conditions:

**(i)** $\operatorname{Jac}(p,q):=p_xq_y-p_yq_x \in \mathbb{C}-\{0\}$.

(Generally, $\operatorname{Jac}(p,q) \in \mathbb{C}(x,y)$).

**(ii)** One of $\{p,q\}$ can be written as $\frac{u}{v}$, where $u,v \in \mathbb{C}[x,y]$, $\gcd(u,v)=1$ and (exactly) one of $\{u,v\}$ is a multiple of $y$.

(Edit: The original condition (ii) was slightly different and unclear).

Question:Is such $f$ necessarily an automorphism of $\mathbb{C}(x,y)$?

**Examples:** $f: (x,y) \mapsto (xy^2,\frac{1}{y})$. We have, $\operatorname{Jac}(xy^2,\frac{1}{y})=-1$.
It is clear that $f$ is an automorphism of $\mathbb{C}(x,y)$ (obviously, $x$ and $y$ are in the image of $f$).

$g: (x,y) \mapsto (x^2,y^2)$ is not an automorphism of $\mathbb{C}(x,y)$, but this does not contradict a positive answer to my question, since $g$ satisfies condition (ii) but does not satisfy condition (i).

**Motivation:** If we replace $\mathbb{C}(x,y)$ by $\mathbb{C}[x,y]$, then by a known result concerning the Newton polygon we obtain that such $f$ is an automorphism of $\mathbb{C}[x,y]$.

The known result can be found, for example, in Essen's book Proposition 10.2.6, in Cheng-Wang's paper Lemma 1.14, Nowicki-Nakai's paper Proposition 2.1 and Nagata's paper.

**Remark:**
I suspect that the answer to my above question is yes, but I am not sure if the proof for $\mathbb{C}[x,y]$ can be adjusted to $\mathbb{C}(x,y)$.

Thank you very much!

everyelement is a multiple of $y$ and of $1/y$. $\endgroup$ – LSpice May 18 at 1:33