That was supposed to be C subscript 6, etc.
i suppose you mean C6 x C3, or Z6 x Z3, to which the former is isomorphic. this group has order 18 = (6)(3), so the only possible orders are 1,2,3,6,9 and 18.
suppose (a,b) is any arbitrary element of C6 x C3. then (a,b)^6 = (a^6,b^6) = (e_C6,e_C3), so we don't have any elements of orders 9 or 18.
it should be clear that (a,b) is of order 6 iff a is of order 6, or if a is of order 2, and b is of order 3.
then the elements of order 6 are: (x,e), (x,y), (x,y^2), (x^5,e), (x^5,y), (x^5,y^2), (x^3,y), (x^3,y^2)
it should also be clear that (a,b) is of order 3 if a is of order 3, or if a = e_C6, and b ≠ e_C3.
so the elements of order 3 are: (x^2,e), (x^2,y), (x^2,y^2), (x^4,e), (x^4,y), (x^4,y^2), (e,y), (e,y^2).
if (a,b) is of order 2, we have to have b = e_C3, since 2 does not divide 3, and a of order 2, which means the ONLY element
of order 2 is (x^3,e). that's 17 elements, and of course (e,e) is of order 1.