I am interested in the question above. I know that the answer is NO if the base field is for instance $\mathbb{Q}$ (the intersection of $\mathbb{Q}(x^2)$ and $\mathbb{Q}((x-1)^2)$ is $\mathbb{Q}$ where $x$ is a $\mathbb{Q}$-transendental element).

But how about two function fields over a finite field $\mathbb{F}_q$?

Let $F$ be a function field over $\mathbb{F}_q$ and $F_1$ and $F_2$ two sub function fields of $F$ such that both extensions are separable. The counterexample from above doesn't work here. More generally: If $F/F_1$ and $F/F_2$ are Galois with the corresponding subgroups $U_i$ of the finite (here one needs a finite field) automorphism group $G$ of $F$, then, by finite Galois theory, $F_1 \cap F_2$ is the fixed field of the subgroup which is generated by $U_1$ and $U_2$. Since the subgroup is finite, the fixed field must be a function field.

But, what if only one of the extensions is Galois (the other case (both non-Galois) should follow from this case by considering the Galois closure of one of the extensions)? Or is the claim not true and somebody can provude a counterexample?

Ansatz 1: Is it possible to construct an extension $E/F$ such that $E/F_1$ and $E/F_2$ are Galois?

Ansatz 2: Starting with the easiest case: Let $F=\mathbb{F}_q(x)$ be a rational function field and $F_i = \mathbb{F}_q(f_i(x))$ with polynomials $f_i$. It boils down to the question whether there are polymomials $g_i$ such that $$ g_1(f_1(x)) = g_2(f_2(x)). $$

I tried out some specific examples, but couldn't find an extendable pattern. For $q=3$ and $f_1 = x^2$ and $f_2 = (x-1)^2$ (example from above), we can choose the $g_i$ in the following way $$ (x^2)^3 + (x^2)^2 + x^2 = x^6 + x^4 + x^2 = ((x-1)^2)^3 + ((x-1)^2)^2 + (x-1)^2. $$

Thank you.