Below is a quite verbose expansion of @LordSharktheUnknown's answer. Though this appears not suited for a beginner in commutative algebra.
$\def\f{\Bbb F}\def\z{\mathbf{Z}}$
We want to find all prime ideals of ${\Bbb F}_3[x,y]/(y^2-x^3+x)$ whose
intersection with $\f_3[x]$ is $(x^2+x+2)$.
The elliptic curve $y^2=x^3-x$ means the scheme
$\text{Spec}(\f_3[x,y]/(y^2-x^3+x))$. Denote it as $E$. Then
the prime ideals in the quotient ring are nothing but points of $E$
over $\f_3$. Denote the residue field of $p$ as $\kappa_p$, i.e.
$$\kappa_p=\f_3[x,y]/p.$$
Now consider the maps:
$$\f_3[x]\rightarrow\f_3[x,y]\rightarrow\f_3[x,y]/(y^2-x^3+x).$$
And for every point $p$ on $E$ over $\f_3$, consider the composite map:
$$\f_3[x]\rightarrow\f_3[x,y]/(y^2-x^3+x)\rightarrow\f_3[x,y]/p=\kappa_p.$$
If the intersection of $p$ with $\f_3[x]$ is $(x^2+x+2)$, then this
map has kernel $(x^2+x+2)$, and hence induces the map
$$\f_3[x]/(x^2+x+2)\hookrightarrow\kappa_p.$$
But this ring $\f_3[x]/(x^2+x+2)$ is isomorphic to $\f_9$, as the polynomial
$x^2+x+2$ is irreducible over $\f_3$, thus the residue field $\kappa_p$ is an
extension of $\f_9$. That is to say, $p$ is a point of $E$ defined over $\f_9$
if and only if the intersection of $p$ with $\f_3[x]$ is $(x^2+x+2)$.
It is a standard fact that the ring $\f_3[x,y]/(y^2-x^3+x)$ is a Dedekind domain, so every
ideal can be uniquely factored into products of prime ideals. Therefore we are
looking for the prime ideal factorisation of the ideal $(x^2+x+2)$.
By the fact that every ideal in a Dedekind domain is generated by at most two
elements, we can write $p$ as $(x^2+x+2,f(x,y))$ for some polynomial $f$.
Now $p$ contains $(y^2-x^3+x)$ means that $y-x^3+x$ can be written as a linear
combination of $x^2+x+2$ and $f$: $y-x^3+x=\alpha(x^2+x+2)+\beta f$ for some
$\alpha$ and $\beta$.
A calculation shows that $x^3-x=(x+1)^2\pmod{x^2+x+2}$. So we find that $f=y-x-1$ and $f=y+x+1$ satisfy the condition, and they give all the prime ideals we want to find.
Hope this helps.
\mathbb{F}_3
in your formulas (assuming the formula as a whole is enclosed in dollar signs). – zipirovich Aug 13 '18 at 02:25