First the number of cyclic subgroups of order 9: If H=<(a,b,c)> has order 9, then c has order 9. So with an appropriate "power", we may take c=1. Further 2(a,b,c) is in H. So there are really just 2 choices each for a and b, giving a total of 4 cyclic subgroups of order 9.
Next if H is not cyclic of order 9, then it is an order 9 subgroup (subspace) of the 3 dimensional vector space over the field with 3 elements. The total number of sets of 2 independent vectors is (3^3-1)(3^3-3)=26*24. Each subgroup has (3^2-1)(3^2-3)=48 sets of 2 independent vectors. Now different subgroups can not have a set of 2 independent vectors in common. So if n is the number of non-cyclic subgroups of order 9, 48n=26*24 or n=13.
Thus finally there are 4+13=17 subgroups of order 9.