Hint: the roots of $\,x^3-3\,$ over $\,\Bbb Q\,$ are $\,\sqrt[3] 3\,,\,w\sqrt[3] 3\,,\,w^2\sqrt[3] 3\,\,\,,\,\,w=e^{2\pi i/3}\,$, thus you have the following subfields:
$$\Bbb Q\subset \Bbb Q(\sqrt[3] 3)\,,\,\Bbb Q(w)\,,\,\Bbb Q(w\sqrt[3] 3)\,...etc.$$
Try to take it from here (Check the automorphism, what each of them fixes, etc.)
Added on request: Well, you're supposed to know already that
$$\Bbb Q\subset\Bbb Q(\sqrt[3] 3)\subset\Bbb Q(\sqrt[3] 3\,,\,w)=:L\,\,,\,[L:\Bbb Q]=6$$
and since $\,\left\{1,3^{1/3}3^{2/3}\right\}\,$ is a basis for $\,\Bbb Q(\sqrt[3]3)/\Bbb Q\,$ and $\,\{1,w\}\,$ is a basis for $\,L/\Bbb Q(\sqrt[3]3)\,$ (why?) , then $\,\left\{1,\,3^{1/3},\,3^{2/3},\,w,\,3^{1/3}w,\,3^{2/3}w\right\}\,$ is a basis for $\,L/\Bbb Q\,$.
So let us check the action of each automorphism in $\,G:=Gal(L/\Bbb Q)\cong S_3\,$ on the above basis, for example: naming the roots of $\,x^3-3\,$ as $\,a_1=\sqrt[3]3\,,\,a_2=\sqrt[3]3\,\,w\,,\,a_3=\sqrt[3]3\,\,w^2\,$ , and taking the automorphism $\,\sigma\in G\,$ that we identify with the involution $\,(1,2)\in S_3\,$ we get:
$$\sigma a_1=a_2\,\,\,,\,\,,\sigma a_2=a_1\,\,\,,\,\,\sigma a_3=a_3$$ so
$$\sigma\left(\begin{cases}1\\{}\\3^{1/3}\\{}\\3^{2/3}\\{}\\w\\{}\\3^{1/3}\,\,w\\{}\\3^{2/3}\,\,w\end{cases}\right)=\begin{cases}1\\{}\\3^{1/3}\,\,w\\{}\\-3^{2/3}\,\,w-3^{2/3}=3^{2/3}\,\,w^2\\{}\\w^2\\{}\\3^{1/3}\\{}\\-3^{2/3}-3^{2/3}\,\,w^2=3^{2/3}\,\,w\end{cases}$$
We can see that $\,\sigma\Bbb Q(3^{1/3}\,w)=\Bbb Q(3^{1/3}\,w)\,$ , and we can even check by means of the Fundamental Theorem of Galois Theory:
$$3=[\Bbb Q(3^{1/3}\,w):\Bbb Q]=[G:\langle \sigma \rangle]$$
and, btw, $\,\langle\sigma\rangle\ntriangleleft G\,\Longleftrightarrow \Bbb Q(3^{1/3}\,w)/\Bbb Q$ not a normal extension.
