The following question is a direct continuation of this elaborate question; it is mentioned there at the end:

Let $u,v \in \mathbb{C}(x,y)$ or $u,v \in \mathbb{C}[x,y]$, if it is easier to answer in this case.

Assume that the following condition, call it $D(f)$, is satisfied for every $f \in \mathbb{C}[x]-\mathbb{C}$:

$\mathbb{C}(u,v,f)=\mathbb{C}(x,y)$.

Question:Is it true that $\mathbb{C}(u,v)=\mathbb{C}(x,y)$?

The notation is of fields of fractions.

Remarks:

- It seems that $y \in \mathbb{C}(u,v)$, but actually I am not sure about this, since I only have the following argument:

We know that $y \in \mathbb{C}(u,v,f)$, so there exist $F,G \in \mathbb{C}[r,s,t]$ such that $y=\frac{F(u,v,f)}{G(u,v,f)}$.

For some $\alpha \in \mathbb{C}$, $f(\alpha)=0$, and then $y=\frac{F(u(\alpha,y),v(\alpha,y),f(\alpha))}{G(u(\alpha,y),v(\alpha,y),f(\alpha))}= \frac{F(u(\alpha,y),v(\alpha,y))}{G(u(\alpha,y),v(\alpha,y))}$, which just shows that $y \in \mathbb{C}(u(\alpha,y),v(\alpha,y))$.

I do not know how to show that $x \in \mathbb{C}(u,v)$, if this is true.

I guess such $u,v$ must be algebraically independent.

Thank you very much!