I understand the definition of a primitive root of integers; however, I am quite confused trying to find the primitive roots of $\frac{\mathbb{Z}/3\mathbb{Z}[x]}{(x^3-x+\overline{1})}$. I know that $\frac{\mathbb{Z}/3\mathbb{Z}[x]}{(x^3-x+\overline{1})}$ has elements of the form $ax^2+bx+c$, with $a,b,c \in \mathbb{Z}/3\mathbb{Z}[x]$, so it has $27$ elements, and $\mathbb{F}_{27} \simeq \frac{\mathbb{Z}/3\mathbb{Z}[x]}{(x^3-x+\overline{1})}$.
Also, it is not clear to me if I must find an element of order 26 or an element of order $\varphi(27) = 18$, where $\varphi$ is the Euler Totient Function.
Could someone give me a tip?
As Leoli1 commented, you have to find an element of order $26$,
since the multiplicative group of $\mathbb F_{27}$ has order $26$.
You may consider this brute force, but using $x^3\equiv x-1$ we have
$x^4\equiv x^2-x, $ $x^5\equiv -x^2+x-1$, $x^6\equiv x^2+x+1$, $x^7\equiv x^2+2x+2$, $x^8\equiv2x^2-1$,
$x^9\equiv x+1, x^{10}\equiv x^2+x$, $x^{11}\equiv x^2+x-1$, $x^{12}\equiv x^2-1$, and $x^{13}\equiv-1$.
It follows that $x$ is a primitive root.
(I could have taken a short cut: $x^{12}\equiv (x^4)^3\equiv (x^2-x)^3\equiv(x^2)^3-x^3= x^6-x^3$
$=x^3(x^3-1)\equiv (x-1)(x-2)\equiv (x-1)(x+1)=x^2-1$.)