I know that the Galois group $f(t)=(t^3-2)(t^3-3)$ over $\mathbb{Q}$ is $Gal(\mathbb{Q}(2^{\frac{1}{3}},3^{\frac{1}{3}},\xi)/\mathbb{Q})$ (where $\xi$ is the 3rd root of unity) and its order is 18. I also know that this group should be some non-abelian subgroup of $S_3 \times S_3$. (here, $S_n$ is the symmetric group)
But I don't know the exact shape of this Galois group.
How can we know this group?
Given that the Galois group acts on the roots this gives us intution for defining automorphisms: let $\sigma, \tau, \eta : \mathbb{Q}(2^{1/3},3^{1/3},\xi) \to \mathbb{Q}(2^{1/3},3^{1/3},\xi)$, $\sigma(2^{1/3}) = \xi2^{1/3}$, $\tau(3^{1/3}) = \xi3^{1/3}$, $\eta(\xi) = \xi^2$. By composing such maps we generate a group of order $18$ so this must be the whole group. Its easy to see this group is not abelian (as you are aware).
$\sigma$ and $\tau$ generate an abelian subgroup of order $9$, which you know will be normal (index $2$). You can construct an exact sequence $$\langle \sigma, \tau \rangle \to \textrm{Gal}(\mathbb{Q}(2^{1/3},3^{1/3},\xi)/\mathbb{Q}) \to \langle \eta \rangle$$ and from here you'll be able to exhibit the Galois group as a direct product or a semi-direct product (hint: direct product of abeian groups is abelian).
Here's a more mechanical approach. By the Sylow theorems and other tricks from Group theory we can identify all non-abelian groups of order 18: $$D_{18}, \hspace{1mm} S_3 \times \mathbb{Z}/3\mathbb{Z}, \hspace{1mm} (\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}) \rtimes \mathbb{Z}/2\mathbb{Z}$$ Now look at the orders of the elements in each group and it should be clear which group you can identify the Galois group with.