Here is one way you might proceed: we know that a matrix $A$ will be invertible if and only if its columns form a basis for $\Bbb Z_3 \times \Bbb Z_3$.
So that means that we can choose any non-zero vector $(a,b)$ for the first column (here, it might be more profitable to view $\Bbb Z_3$ as the congruence classes of $-1,0,1$, instead of $0,1,2$).
That means we have 8 choices for our first vector.
Now the second vector will form a linearly dependent set with the first if (and only if):
a) The second vector is 0
b) the second vector is the same vector as the first
c) the second vector is the negative of the first
So for each of the possible 8 vectors we chose, we have 6 possible choices for the 2nd vector, giving 48 in all.
You should, knowing this, be able to explicitly write down all 48 matrices. Why must exactly half of these have determinant 1?