To prove that $13$ is a Wilson prime, you need to show that
$$12! \equiv -1 \pmod {13^2} \,.$$
To make the computations faster, observe that
$$12!= 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot (2 \cdot 5)\cdot 11 \cdot (4 \cdot 3)$$
We split the product in three:
$$2 \cdot 3 \cdot 4 \cdot 7 =(2 \cdot 7)(3 \cdot 4)=(13+1)(13-1)\equiv -1 \pmod {13^2} $$
Also
$$6 \cdot 11 \cdot 8 \cdot 2 \cdot 4 = (6 \cdot 11)(8 \cdot 8)=(5 \cdot 13+1)(5 \cdot 13-1) \equiv -1 \pmod{13^2}$$
and
$$ 5 \cdot 9 \cdot \cdot 5 \cdot 3=25 \cdot 27=(2 \cdot 13-1)(2 \cdot 13+1) \equiv -1 \pmod {13^2} \,.$$
Multiplying we get
$$12! \equiv -1 \pmod{13^2} \,.$$