The comment from Matho123 proves that $\alpha \beta=3$ and that the Galois group of $\mathbb Q(\alpha)$ is normal.
Regarding the Galois group $G$ of $\mathbb Q(\alpha)$, what are its elements? You have the identity $\sigma_1$.
Consider $\sigma_2$ such that $\sigma_2(\alpha)=-\alpha$. Then $$3=\sigma_2(3)=\sigma_2(\alpha \beta)=\sigma_2(\alpha) \sigma_2(\beta)=-\alpha \sigma_2(\beta)$$ Hence $\sigma_2(\beta)=-\beta$ and the order of $\sigma_2$ is equal to $2$.
By similar considerations, one can prove that the elements $\sigma_3,\sigma_4$ of the Galois group $G$ such that $\sigma_3(\alpha)=\beta$, $\sigma_4(\alpha)=-\beta$ are also of order $2$.
Finally, the Galois group has four elements and all elements (except the identity) have order $2$. So $G$ is isomorphic to the Klein four-group.