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

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

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

(ii) Considered one of ${p,q}$ is a a number of of $y$ or a a number of of $frac{1}{y}$.

Query: Is such $f$ essentially an automorphism of $mathbb{C}(x,y)$?

Examples: $f: (x,y) mapsto (xy^2,frac{1}{y})$. We now have, $operatorname{Jac}(xy^2,frac{1}{y})=-1$.
It’s clear that $f$ is an automorphism of $mathbb{C}(x,y)$ (clearly, $x$ and $y$ are within the picture of $f$).

$g: (x,y) mapsto (x^2,y^2)$ isn’t an automorphism of $mathbb{C}(x,y)$, however this doesn’t contradict a optimistic reply to my query, since $g$ satisfies situation (ii) however doesn’t fulfill situation (i).

Motivation: If we change $mathbb{C}(x,y)$ by $mathbb{C}[x,y]$, then by a identified end result regarding the Newton polygon we acquire that such $f$ is an automorphism of $mathbb{C}[x,y]$.

Comment:
I think that the reply to my above query is sure, however I’m not positive if the proof for $mathbb{C}[x,y]$ could be adjusted to $mathbb{C}(x,y)$.

Thanks very a lot!