Question:
What is the Galois group of $f(x)=(x^3-2)(x^3-3)$ over $\mathbb Q$, and what are the subfields which contain $\mathbb Q(\zeta_3)$?
The roots of $f(x)$ are $\sqrt[3]{2},\sqrt[3]{2}\zeta_3,\sqrt[3]{2}\zeta_3^2$ and $\sqrt[3]{3},\sqrt[3]{3}\zeta_3,\sqrt[3]{3}\zeta_3^2$.
Then $E=\mathbb Q[\sqrt[3]{2},\sqrt[3]{3},\zeta_3]$ is the splitting of the separable polynomial $f(x)$ over $\mathbb Q$.
$[E:Q]=18=\operatorname{Gal}(E/\mathbb Q)$
Let $$\begin{align*} \sigma_1:\sqrt[3]{2} &\rightarrow \sqrt[3]{2}\zeta_3\\
\sqrt[3]{3} &\rightarrow \sqrt[3]{3}\\
\zeta_3 &\rightarrow \zeta_3 \end{align*}$$
Then for $\langle \sigma_1 \rangle$ we have the fixed field $\mathbb Q(\zeta_3,\sqrt[3]{3})$ which contains $\mathbb Q(\zeta_3)$.
Similarly if $$\begin{align*} \sigma_2:\sqrt[3]{2} &\rightarrow \sqrt[3]{2}\\
\sqrt[3]{3} &\rightarrow \sqrt[3]{3}\zeta_3\\
\zeta_3 &\rightarrow \zeta_3 \end{align*}$$
then for $\langle\sigma_2\rangle$ we have the fixed field $\mathbb Q(\zeta_3,\sqrt[3]{2})$ which contains $\mathbb Q(\zeta_3)$.
Let $\{1,\sigma_1,\sigma_1^2\}=\langle \sigma_1\rangle=H_1$, $\{1,\sigma_2,\sigma_2^2\}=\langle\sigma_2\rangle=H_2$, $H_3=\langle\sigma_1\sigma_2\rangle$, $H_4=\langle\sigma_1^2\sigma_2\rangle$.
Then the subgroup
$H=\{1,\sigma_1,\sigma_1^2,\sigma_2,\sigma_2^2,\sigma_1\sigma_2,(\sigma_1\sigma_2)^2,\sigma_1^2\sigma_2,\sigma_2^2\sigma_1 \} \cong \mathbb Z_3\times \mathbb Z_3$
so the fixed fields corresponding to $H_1,H_2,H_3,H_4$ are the subfields that contain $\mathbb Q(\zeta_3)$.
Best Answer
You know that $E=\mathbb{Q}(\zeta_3,\root3\of2,\root3\of3)$ is the splitting field and you know that it is of degree 18 over the rationals. Therefore the Galois group $G=\operatorname{Gal}(E/\mathbb{Q})$ is of order 18.
You can get a handle of $G$ as follows. I set as a goal to describe $G$ as a subgroup of $S_6$, because $G$ acts faithfully on the set of six roots $z_1=\root3\of2, z_2=z_1\zeta_3,z_3=z_1\zeta_3^2,$ $z_4\root3\of3$, $z_5=z_4\zeta_3 $, $z_6=\zeta_3^2z_4$ of $f(x)$. As $f(x)$ is not irreducible, the action will not be transitive, but let's not worry about that.
Let $\sigma\in G$ be arbitrary. You know that $\sigma$ will be fully determined, if we know $\sigma(\root3\of2)$ (three choices), $\sigma(\root3\of3)$ (three choices) and $\sigma(\zeta_3)$ (two choices). Altogether there are $3\cdot3\cdot2=18$ possible combinations for these three images. As there are exactly $18$ automorphisms, we can immediately conclude that all the 18 combinations will yield valid automorphisms. This is not always the case, because sometimes there are 'hidden' relations among the generators of a field extension, and the automorphisms must respect these relations, but this time a counting argument saved the day.
So you can go through all the 18 combinations of possible values of $\sigma(\root3\of2),\sigma(\root3\of3),\sigma(\zeta_3)$, and then calculate the corresponding permutation of the six roots. Of course, once you get a few of those calculated, you can use the fact that $G$ is a subgroup of $S_6$ to get others.
While working on this you will also probably learn (by observation!) ways of identifying sets of generators for $G$.
Finding the intermediate fields containing $\zeta_3$ amounts to finding the fixed fields of the subgroups $K\le H$, where $H$ consists of those nine elements of $G$ that map $\zeta_3$ to itself.
Don't hesitate to ask for more help/hints, if you get stuck at some point.
Realizing $G$ as a subgroup of $S_6$ is not strictly necessary to answer your question, but it is a useful exercise in its own, and having that data at your finger tips does help answering your question also.
Edit (after jim added a description of the group $H\cong C_3\times C_3$, and simultaneously while jim added a description of the subgroups $H_i,i=1,2,3,4.$).
All the non-identity elements of $H$ are of order three. So if you pick any one of them, call it $\tau$, it will (alone) generate a subgroup of order three consisting of $\tau,\tau^2=\tau^{-1}$ and the identity. So clearly $\tau$ and $\tau^{-1}$ will generate the same subgroup. Therefore there are a total of four subgroups of order three - each containing two out of the eight non-identity automorphisms in $H$. You have already identified the fixed fields of $\langle \sigma_1\rangle$ and $\langle \sigma_2\rangle$. What about $\langle \sigma_1\sigma_2\rangle$ and $\langle \sigma_1\sigma_2^2\rangle$?
Hints: The numbers $\root3\of6$, $\root3\of{12}$, $\root3\of{18}$ and $\root3\of{36}$ are all in $E$. Are they fixed points of any of your automorphisms? Will all of them generate distinct subfields?
Added after jim had solved the exercise: These cube roots appear in $\mathbb{Q}(\root3\of2,\root3\of3)$ as products of those real generators much like $\sqrt{6}\in\mathbb{Q}(\sqrt2,\sqrt3)$. Observe that $$ (\root3\of6)^2=\root3\of{36}=\frac{6}{\root3\of6} $$ and $$ \root3\of{18}=\root3\of{\frac{216}{12}}=\frac6{\root3\of{12}}, $$ so any field containing one of these pairs of cubic roots will also contain the other.
With the data in OP it is easy to see that for $\alpha=\sigma_1\sigma_2$ we have $$ \alpha(\root 3\of{12})=(\alpha(\root3\of2))^2\alpha(\root3\of3)=\zeta_3^2(\root3\of2)^2\cdot\zeta_3\root3\of3=\root3\of{12} $$ and for $\beta=\sigma_1^2\sigma_2$ we have $$ \beta(\root 3\of{6})=\beta(\root3\of2)\beta(\root3\of3)=\zeta_3^2(\root3\of2)\cdot\zeta_3\root3\of3=\root3\of{6}. $$ Therefore $\operatorname{Inv}(\langle\alpha\rangle)=\mathbb{Q}(\root3\of{12},\zeta_3)$ and $\operatorname{Inv}(\langle\beta\rangle)=\mathbb{Q}(\root3\of{6},\zeta_3)$, as Galois correspondence tells us that the fixed fields are sextic.