Denote the roots of $x^{3} - 2$ by $\theta_{1}, \theta_{2}, \theta_{3}$. Define the map as $$ \begin{equation} \mathrm{Aut}\left( \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q} \right) \to S_{3}, \qquad \sigma \mapsto \tau_{\sigma} \end{equation} $$ such that for $\sigma \in \mathrm{Aut}\left( \mathbb{Q}(\sqrt{ 2 }, \sqrt{ -3 }) / \mathbb{Q} \right)$, the permutation $\tau_{\sigma} \in S_{3}$ is defined as $\tau_{\sigma}(i) = j$ if and only if $\sigma(\theta_{i}) = \theta_{j}$.

First we need to check that the map $(1)$ is indeed a group homomorphism. Then simply noticing that $\theta_{1}, \theta_{2}, \theta_{3}$ generate $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$ over $\mathbb{Q}$ since $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$ is the splitting field of $x^{3} - 2$ can show that $(1)$ is injective. To show the surjectivity, it remains to prove that every $\tau \in S_{3}$ corresponds to a valid $\sigma \in \mathrm{Aut}\left( \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q} \right)$ such that $\tau = \sigma_{\tau}$.