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](https://books.google.co.il/books/about/Polynomial_Automorphisms.html?id=wKJqqd8t8KUC&redir_esc=y) Proposition 10.2.6, in [Cheng-Wang's paper](https://link.springer.com/chapter/10.1007/978-94-015-8555-2_10) Lemma 1.14, [Nowicki-Nakai's paper](https://www.researchgate.net/publication/265368034_On_Appelgate-Onishi's_Lemmas) Proposition 2.1 and [Nagata's paper](https://core.ac.uk/download/pdf/39334102.pdf).


**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!



Certain endomorphisms of $\mathbb{C}(x,y)$
The answer is **no**.


Take $p=\frac{x^2}{2}$, take $q=\frac{y}{x}$. The Jacobian matrix is $\begin{pmatrix} x& -\frac{y}{x^2} \\ 0 & \frac{1}{x}\end{pmatrix}$ whose determinant is equal to $1$. However, $f$ is definitely not an automorphism of $\mathbb{C}(x,y)$.


More generally, take any polynomial $p\in\mathbb{C}[x]$ and choose $q=\frac{y}{p\_x}$. This gives you a counterexample as soon as $\deg(p)\ge 2$.


