Let $f=f(x,y),g=g(x,y) \in \mathbb{C}[x,y]$, each of degree $\geq 1$, and $f,g$ are algebraically independent over $\mathbb{C}$ (= their Jacobian $\in \mathbb{C}[x,y]-\{0\}$).

(1)Is there a sufficient condition that will guarantee that $\mathbb{C}(f,g)=\mathbb{C}(x,y)$?

Perhaps it would help if we will consider $f$ and $g$ as polynomials in one variable $y$ over $\mathbb{C}(x) \subset \bar{\mathbb{C}(x)}$?

A related question I have asked is this question, but I have not got an answer for it, so I decided to ask the above related question here, hoping to get some additional help.

**A remark:** I have just found three relevant papers: 1, 2 and
3, where the last one seems to answer my question (Theorem 1.10, or more accurately, since here the base field is of characteristic zero, Theorem 1.10 in characteristic zero is a result of K.P. Russell "Field generators in two variables", J. Math. Kyoto Univ., 1.

(2)Does it make a difference if we allow $f,g \in \mathbb{C}(x,y)$?

For example, $f=x$ and $g=\frac{1}{y}$ (what is the analog notion of being algebraically independent?).

Any comments are welcome! Thank you.