Let $u$ be the real root of $x^3-x+1$; I am trying to calculate de Galois group of $\mathbb{Z}/3\mathbb{Z}(u)/\mathbb{Z}/3\mathbb{Z}$; let's start with the first one; as $x^3-x+1$ is monic, and irreducible over $\mathbb{Z}/3\mathbb{Z}$ (just checking that $0$, $1$ and $2$ are not roots of the polynomial), and $u$ is root of the polynomial, $x^3-x+1$ is the minimal polynomial of $u$ over $\mathbb{Z}/3\mathbb{Z}$. So, $[\mathbb{Z}/3\mathbb{Z}(u):\mathbb{Z}/3\mathbb{Z}]=3$ (the degree of the extension), being possible to choose $\{1,u,u^{2}\}$ as a basis for the extension. Then, we proceed to calculate the Galois group, i.e., $G=Aut_{\mathbb{Z}/3\mathbb{Z}}\mathbb{Z}/3\mathbb{Z}(u)$ (the group of $\mathbb{Z}/3\mathbb{Z}$-automorphisms of $\mathbb{Z}/3\mathbb{Z}(u)$). We can conclude that we can define every automorphism by the image of $u$, and if $\varphi \in G$, $\varphi(u)$ is root of the minimal polynomial of $u$ ($x^3-x+1$); of course, some minor proofs have to be given in order to justify this assumptions. The only root of the polynomial that is in $\mathbb{Z}/3\mathbb{Z}(u)$ is $u$, as the other $2$ are complex roots, so the only automorphism possible is the identity, that sends $u$ to $u$. So, $G=\{Id\}$, and $G'$ (i.e, ''what the automorphisms on $G$ fix from $\mathbb{Z}/3\mathbb{Z}(u)$ '') is $\mathbb{Z}/3\mathbb{Z}(u)$, and as $G'$ is different from $\mathbb{Z}/3\mathbb{Z}$, the extension is not a Galois extension. Is this sketch of my way of thinking the exercise ok? And, If I change $\mathbb{Z}/3\mathbb{Z}(u)$ with any $\mathbb{Z}/n\mathbb{Z}(u)$, with $n$ prime, does the exercise will still have analogous results? Thanks in advance for your time!
Important remark Every time I write $\mathbb{Z}/3\mathbb{Z}(u)/\mathbb{Z}/3\mathbb{Z}$ I am refering to $[(\mathbb{Z}/3\mathbb{Z})(u)]/(\mathbb{Z}/3\mathbb{Z})$, i.e., the extension $(\mathbb{Z}/3\mathbb{Z})(u)$ over $\mathbb{Z}/3\mathbb{Z}$
In your example the other two roots of $x^3-x+1$ lie in the field $\mathbb{Z}/3\mathbb{Z}(u)$. Explicitly the three roots are $u,u+1,u-1$. It is easy to see that the automorphism $\varphi\colon u\mapsto u+1$ has order $3$ and thus generates the Galois group.
In general, given an irreducible polynomial of degree $d$ over a finite field $GF(p^k)$, if we attach a single root $u$, we obtain $GF(p^{kd})$. This sits inside the splitting field of the polynomial. However any other root generates a subfield of the same size, and as there is a unique subfield of size $p^{kd}$ (namely the set of solutions to $x^{p^{kd}}=x$), all roots generate the same subfield.
We conclude that $GF(p^{kd})$ is already a splitting field for the polynomial.