Let $k$ be a field of characteristic zero, for example $k=\mathbb{R}$ or $k=\mathbb{C}$.
Of course, $k(x^2,x^3)=k(x)$, since $x=\frac{x^3}{x^2}$.
Let $f_1,\ldots,f_n,g_1,\ldots,g_m \in k[x]$, $n,m \geq 1$.
Denote: $F(T)=f_nT^n+\cdots+f_1T+x^2$ and $G(T)=g_mT^m+\cdots+g_1T+x^3$.
Let $h \in k[x]$.
Is it possible to characterize $f_1,\ldots,f_n,g_1,\ldots,g_m \in k[x]$, $n,m \geq 1$, such that for every $h \in k[x]$ we have: $k(F(h),G(h))=k(x)$?
I could not find such $f_i,g_j,n,m$.
For example: $F(T)=T+x^2$, $G(T)=T+x^3$. For $h=x^3$ we get $F(h)=x^3+x^2$, $G(h)=x^3+x^3=2x^3$, so clearly, $k(F(h),G(h))=k(x)$. For $h=x^4-x^3$ we get $F(h)=x^4-x^3+x^2$, $G(h)=x^4-x^3+x^3=x^4$, abd then, $k(F(h),G(h)) \subsetneq k(x)$.
I think that for arbitrary fixed $f_i,g_j,n,m$, there exists $h \in k[x]$ such that $k(F(h),G(h)) \subsetneq k(x)$, but not sure how to prove this.
I am familiar with D-resultant, but it seems difficult to apply it. The theorem relevant here is that $D-\operatorname{Res_s}(\frac{(F(h))(s)-F(h)}{s-x},\frac{(G(h))(s)-G(h)}{s-x})$ is a non-zero element of $k[x]$ if and only if $k(F(h),G(h))=k(x)$. See this paper.
If we apply this to $x^2,x^3$ we obtain that their D-resultant is $x^2$, and indeed $k(x^2,x^3)=k(x)$.
Also, I wonder if algebraic geometry could help solve my question.
Any comments are welcome; thank you!
Edit: (1) I do not mind to further assume that for every $h \in k$, $k(F(h),G(h)) = k(x)$. In the above example, this is satisfied. (2) $f_1,\ldots,f_n,g_1,\ldots,g_m \in k[x]$, not necessarily in $k$.
Now asked in MO.