Let $f=f(t), g=g(t) \in \mathbb{C}[t]$.
Assume that:
(1) $\deg(f) \geq 2,\deg(g) \geq 2$.
(2) $\mathbb{C}(f(t),g(t))=\mathbb{C}(t)$.
(3) The set of zeros of $f$, $S_f$, equals the set of zeros of $g$, $S_g$; denote $S_f=S_g=\{c_1,\ldots,c_m\}$.
(4) $\langle f'(t),g'(t) \rangle = \mathbb{C}[t]$ (namely, the ideal generated by $f'$ and $g'$ is the unit ideal).
Now, (4) implies that $f'$ and $g'$ do not have a common zero (there is no need to apply Hilbert's weak Nullstellensatz), so from (3) we obtain that:
$f(t)=\lambda(t-c_1)^{d_1}\cdots(t-c_l)^{d_l}(t-c_{l+1})^{d_{l+1}}\cdots(t-c_m)^{d_m}$,
$g(t)=\mu(t-c_1)^{e_1}\cdots(t-c_l)^{e_l}(t-c_{l+1})^{e_{l+1}}\cdots(t-c_m)^{e_m}$,
where $\lambda, \mu \in \mathbb{C}^{\times}$,
and $d_i,e_i \geq 1$ are such that if $d_i \geq 2$ then $e_i=1$ and vice versa (= if $e_i \geq 2$ then $d_i=1$).
The reason is as follows: If, for example, $d_1 \geq 2$ then $c_1$ is also a zero of $f'(t)$, and if $e_1 \geq 2$ then $c_1$ would also be a zero of $g'(t)$, but then $c_1$ would be a common zero of $f'(t)$ and $g'(t)$, contrary to assumption (4).
Is it true that necessarily $\mathbb{C}[f(t),g(t)]=\mathbb{C}[t]$? Or is there a counterexample?
Notice that $f(t)=t^2$, $g(t)=t^3$ is not a counterexample, since conditions (1), (2), (3) are satisfied, but condition (4) is not satisfied.
See also this question.
Edit: After receivig a counterexample, do you think that there exists an additional (not too strong) condition that will guarantee that $\mathbb{C}[f(t),g(t)]=\mathbb{C}[t]$? There is a nice condition which says that $f'(t),g'(t) \in \mathbb{C}[f(t),g(t)]$, so the fifth condition, together with the former four conditions, should imply that $f'(t),g'(t) \in \mathbb{C}[f,g]$.
A 'plausible' fifth condition: (5) $f(t)$ and $g(t)$ are separable polynomials. In that case we must have: $d_i=e_i=1$ for every $1 \leq i \leq m$. Therefore, $f(t)=g(t)$, so from (2) we get that $\mathbb{C}(f(t))= \mathbb{C}(f(t),g(t))=\mathbb{C}(t)$, hence $\deg(f)=1$ which contradicts condition (1)... Therefore, if we omit condition (1) and add condition (5), we obtain that $\mathbb{C}[f(t),g(t)]=\mathbb{C}[t]$.
Thank you very much!