Here is one way to do it that requires knowing only a primitive root $\bmod p$ to get all primitive roots $\bmod p^2$, provided that $p$ is twice a smaller prime plus $1$, so that all nonzero residues besides $\pm 1$ are either quadratic or primitive. I will demonstrate it for $p=5$ letting you apply it for $p=11$.
Start with $2$ as a primitive root $\bmod 5$. Then there will be a series of primitive roots $\equiv 5k+2 \bmod 25$. For one residue $k \bmod 5$ we will have $(5k+2)^4\equiv 1 \bmod 25$ and that value of $k$ must be rejected, but other values of $k$ will all work.
Use the Binomial Theorem to get $(5k+2)^4$, but retain only the first ad zero powers of $k$ (why?). Thus
$(5k+2)^4 \equiv (4)(5k)(2^3)+2^4 \equiv 1 \bmod 25$
$10k+16 \equiv 1, 10k \equiv 10 \bmod 25$
Divide by $10$ noting that the modulus is to be divided by $gcd(10,25)=5$. Then $k\equiv 1 \bmod 5$ meaning $7$ will not be a primitive root $\bmod 25$ but all other residues two greater than a multiple of $5$ will work. We therefore get a subset of primitive roots:
$\{2,12,17,22\}$
Now take the root we rejected, $7$. If we multiply the above primitive roots by it, we get nonprimitive roots because these are quadratic residues. But multiply by $7$ twice, that is by $7^2$, and we get another set of primitive roots because the products are non-quadratic and also miss being $\equiv -1 \bmod 5$. Since $7^2 \equiv 24 \equiv -1 \bmod 25$ this gives another subset of primitive roots
$\{3,8,13,23\}$
Multiplying by $7^2$ again cycles us back to the first subset, so the complete set of primitive roots $\bmod 25$ will be the union of the two subsets
$\{2,3,8,12,13,17,22,23\}$.
Now apply this method to$11=2×5+1$ and see what happens. Since $11$, like $5$, is not one greater or one less than a multiple of $8$, $2$ will be a primitive root $\bmod 11$. Since there are four such primitive roots $\bmod 11$ overall you will need to generate four subsets before taking their union.
Mod $71^2$ we have:
$7^2 \equiv 49$
$7^4 \equiv 49^2 = 2401$
$7^8 \equiv 2401^2 \equiv 2938$
$7^{16} \equiv 2938^2 \equiv 1652$
$7^{32} \equiv 1652^2 \equiv 1923$
$7^{64} \equiv 1923^2 \equiv 2876$
$7^{70} = 7^{64} \cdot 7^{4} \cdot 7^{2} \equiv 2876 \cdot 2401 \cdot 49 \equiv 1563 \not\equiv 1$
This method of finding powers is called exponentiation by squaring.
However, you don't need to compute any of this: $7$ is the smallest primitive root mod $71$ and the smallest primitive root mod $p$ is a primitive root mod $p^2$ for all odd $p<40487$. See OEIS:A055578.
Best Answer
It helps in that you know in advance that the order of a given element must be a divisor of $22$, thus the order must be $1,2,11$ or $22$. You know the elements of order $1,2$ so you just have to rule out $11$.
To be specific: we check that $2^{11}\equiv 3^{11}\equiv 1 \pmod {23}$ so neither are primitive roots. However $5^{11}\equiv -1 \pmod {23}$ so $5$ is a primitive root.