There is a general techinque for finding Galois groups. I shall outline them for this actual case:
Verify that your extension is Galois. For example, I claim that our extension is the splitting field of $x^3-2$. Indeed, the roots of $x^3-2$ are $\sqrt[3]{2},\omega\sqrt[3]{2},\omega^2\sqrt[3]{2}$ and clearly $\mathbb{Q}(\sqrt[3]{2},\omega)$ is equal the field generated by these roots over $\mathbb{Q}$. Thus, our extension is the splitting field of a separable polynomial.
Find the degree of your extension. For us it's pretty simple since one can verify that the following tower holds
$\begin{array}{c}\mathbb{Q}(\sqrt[3]{2},\omega)\\ \vert 2\\ \mathbb{Q}(\sqrt[3]{2})\\ \vert 3\\ \mathbb{Q}\end{array}$
The first from the fact that we know that minimal polynomial $m_{\mathbb{Q},\sqrt[3]{2}}=x^3-2$ and the second since evidently $1+x+x^2\in\mathbb{Q}(\sqrt[3]{2})[x]$ annihilates $\omega$ and $\omega\notin\mathbb{Q}(\sqrt[3]{2})$ (since it's not real). Thus the multiplicative property of towers gives that our extension is degree six.
3.Find how many possible elements can be in the Galois group of your extension. In our case, you have noted that any $\sigma\in\text{Gal}(\mathbb{Q}(\sqrt[3]{2},\omega)/\mathbb{Q})$ must send $\sqrt[3]{2}$ to another root of its minimal polynomial, which there are three (since its minimal polynomial, as we know, is $x^3-2$), and $\omega$ must be sent to a root of its minimal polynomial $1+x+x^2$ of which there are two. Thus, the possible values that $\sigma$ sends $(\sqrt[3]{2},\omega)$ to are $(x,y)$ such that $x\in\{\sqrt[3][2],\omega\sqrt[3]{2},\omega^2\sqrt[3]{2}\}$ and $y\in\{\omega,\omega^2\}$. Since any $\sigma$ is determined by the value it takes on $(\sqrt[3]{2},\omega)$ (since they are generators) you can conclude there are at most six $\sigma$'s (not the statistical framework) and so $|\text{Gal}(\mathbb{Q}(\sqrt[3]{2},\omega)/\mathbb{Q})|\leqslant 6$.
4.Use the fact that our extension is Galois to conclude that $|\text{Gal}(\mathbb{Q}(\sqrt[3]{2},\omega)/\mathbb{Q})|$ is the degree of our extension, which is six. Aha! But we know that our Galois group has at most six elements--those $\sigma$ sending the generators to the six possible choices for $(x,y)$. But, since there MUST be six automorphisms we may conclude that all six possible automorphisms ARE automorphisms. Thus, we have found our Galois group
The obvious trick was to show that even though we can a priori only bound the number of possible automorphisms, using Galois theory we can figure out exactly how many automorphisms we SHOULD have which often forces all possible choices to be all actual choices.
I hope that was helpful. Feel free to ask any questions.