The polynomial $f$ is an irreducible Eisenstein polynomial with $p = 3$.
Its roots are easy to find using the substitution $Y = X^2$ and then the $abc$-formula: $\{\sqrt{\frac{3}{2}+\frac{\sqrt{-3}}{2}}, -\sqrt{\frac{3}{2}+\frac{\sqrt{-3}}{2}}, \sqrt{\frac{3}{2}-\frac{\sqrt{-3}}{2}}, -\sqrt{\frac{3}{2}-\frac{\sqrt{-3}}{2}} \}$.
One automorphism is the complex conjugation, but I have no clue on other maps between these roots.
I think the order of the Galois group is 8. As we have $[\mathbb{Q(\sqrt{\frac{3}{2}+\frac{\sqrt{-3}}{2}})} : \mathbb{Q}] = 4$. Now if I could prove that $\sqrt{\frac{3}{2}-\frac{\sqrt{-3}}{2}}\not \in \mathbb{Q(\sqrt{\frac{3}{2}+\frac{\sqrt{-3}}{2}})}$, then I know that $\sqrt{\frac{3}{2}-\frac{\sqrt{-3}}{2}}$ has a minimal polynomial of degree 2 over $\mathbb{Q(\sqrt{\frac{3}{2}+\frac{\sqrt{-3}}{2}})} $.