We want to compute the Galois group $\mathrm{Gal}(\Bbb{Q}(\omega)/\Bbb Q)$, which is also denoted by $\mathrm{Aut}(\Bbb{Q}(\omega)/\Bbb Q)$, where $\omega:=e^{\frac{2\pi i}{3}}\in \Bbb C$.
We know that $\omega$ is a root of the 3rd cyclotomic polynomial, which is $\Phi_3(X):=X^2+X+1\in \Bbb Z[X]$. Since it is monic, irreducible and with root this element, in fact it is the minimum polynomial of $\omega$ over $\Bbb Q$. Thus, $m_{(\omega,\Bbb Q)}(X)=X^2+X+1\in \Bbb Q[X]$ and the set $\mathcal{B:=\{1,\omega\}}\subseteq \Bbb{Q}(\omega) $ is a basis of the $\Bbb Q$-vector space $\Bbb{Q}(\omega)$.
Let's take now an element $\sigma \in \mathrm{Aut}(\Bbb{Q}(\omega)/\Bbb Q)$. Then, \begin{align*} \sigma : \Bbb{Q}(\omega) & \longrightarrow \Bbb{Q}(\omega),\\ x:=a+b\omega&\longmapsto\sigma(x)=\sigma (a+b\omega)=a+b \sigma( \omega) \end{align*} since $\sigma(\Bbb Q)=\Bbb Q$, by definition. Then, $\omega=e^{\frac{2 \pi i}{3}} \implies \omega^3=1 \implies \sigma( \omega^3) = \sigma(1) \implies \sigma (\omega) ^3=1$ so we can see that $$\sigma(\omega)=e^{\frac{2k\pi i}{3}},\ k=0,1,2.$$ If $k=0\implies \sigma(\omega)=1$, then $\mathrm{Im}\sigma=\{\sigma(a+b):a+b\in \Bbb Q\}=\Bbb Q$, so in this case $\sigma$ is not surjective, and $\sigma \notin \mathrm{Aut}(\Bbb Q(\omega)/\Bbb Q)$.
If $k=1\implies \sigma(\omega)=\omega$, then $\sigma=\mathrm{Id}_{\Bbb Q(\omega)} \in \mathrm{Aut}(\Bbb Q(\omega)/\Bbb Q)$.
If $k=2\implies \sigma(\omega)=\omega^2=-\omega-1$, since $$\Phi(\omega)=0 \iff \omega^2+\omega+1=0 \iff \omega^2=-\omega-1.$$ We can show now that this is a homomorphism from $\Bbb Q(\omega)$ to itself. Also, $$\mathrm{Ker}\sigma=\{a+b\omega:\ \sigma(a+b\omega)=0\}.$$ So, $\sigma(a+b\omega)=0 \iff a+b(-\omega-1)= \iff (a-b)+(-b)\omega=0$ and since $1,\omega$ is a basis, we should have $a-b=0,-b=0\iff a=b=0$. So $\mathrm{Ker}\sigma=\{0\}$ and $\sigma$ is injective. Lastly, \begin{align*} \mathrm{Im}\sigma \quad = & \quad \{\sigma(a+b\omega):\ a+b\omega\in \Bbb Q(\omega)\} \\ \quad = & \quad \{ \underbrace{(a-b)}_{\kappa}+\underbrace{(-b)}_{\lambda}\omega:a+b\omega\in \Bbb Q(\omega) \}.\\ \quad = & \quad \{ \kappa+\lambda\omega: \kappa,\lambda\in \Bbb Q\} \\ \quad = & \quad \Bbb Q(\omega) \end{align*} and so $\sigma$ is surjective and name it $\psi$.
Finally, $\psi \in\mathrm{Aut}(\Bbb Q(\omega)/\Bbb Q)$ and $\mathrm{Aut}(\Bbb Q(\omega)/\Bbb Q)=\{\mathrm{Id}_{\Bbb Q(\omega)},\psi\}\cong (\Bbb Z_2,+)$.
Remark. Another way to work this would be to observe that $\omega=\frac{-1}{2}+i\frac{\sqrt 3}{2}$, so $\Bbb Q(\omega)=\Bbb Q(i\sqrt 3)$.
Could you please verify if everything is correct and well written?
Thank you