Let $f=f(x,y), g=g(x,y) \in \mathbb{C}[x,y]$.
Which $f,g$ satisfy $\mathbb{C}(fy,gy)=\mathbb{C}(x,y)$?
Examples:
(1) $f=2x$, $g=1-2x$. It is easy to see that $\mathbb{C}(2xy,(1-2x)y)=\mathbb{C}(2xy,y-2xy)=\mathbb{C}(2xy,y)=\mathbb{C}(x,y)$.
(2) $f=x$, $g=xy$.
Attempts to generalize the two examples:
(1) $af+bg \in \mathbb{C}[x]-\{0\}$, for some $a,b \in \mathbb{C}$, similarly to example (1). But if $f=x^2$ and $g=x^4$, then this does not work since $\mathbb{C}(x^2y,x^4y)=\mathbb{C}(x^2,y)$. So perhaps we should require $af+bg \in \mathbb{C}-\{0\}$? This will imply that $\mathbb{C}(fy,gy) \ni afy+bgy=y(af+bg)=\lambda y$, and then $f,g \in \mathbb{C}(fy,gy)$, which still does not guarantee that $x \in \mathbb{C}(fy,gy)$.
(2) The following seems to be a false claim: $f$ and $g$ themselves already generate $\mathbb{C}$, namely: $\mathbb{C}(f,g)=\mathbb{C}(x,y)$; although in example (2) it works, but in other cases it may not work.
Edit: After receiving a comment that the resultant may be relevant, I add the following links: Theorem 2.1 and this question.
If I am not wrong, the criterion in the second link implies that, in our case: If $\gcd(f,g)=1$, considering $f,g \in (\mathbb{C}(x))[y]$, then $\mathbb{C}(x)(fy,gy)=(\mathbb{C}(x))(y)=\mathbb{C}(x,y)$.
Thank you very much!