= Solution
First quotient by $3$, obtaining the <quotient ring>
$$
R\cong\mathbb F_3[X]/(X^2+X+1).
$$
<Polynomial division> by this monic quadratic gives a unique representative $a+bX$ with $a,b\in\mathbb F_3$. Uniqueness holds because a nonzero multiple of a monic quadratic cannot have degree less than two. Therefore \b[$|R|=3^2=9$].
In characteristic three, $X^2+X+1=(X-1)^2$. The residue $\epsilon=X-1$ is nonzero by the uniqueness of the degree-one representative, but $\epsilon^2=0$. Thus $\epsilon$ is both a <nilpotent element> and a <zero divisor>. Consequently \b[$R$ is not an integral domain]. Equivalently, its multiplication is that of $\mathbb F_3[\epsilon]/(\epsilon^2)$, rather than that of a field with nine elements.
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