The thing is that an element of order 3 will always generate a subgroup of order 3.

If x has order 3, then x^3 = e, where e is the identity. The group generated by x consists of all powers of x, and is thus {e,x,x^2}. Notice that you forgot the identity in your set, while you included x^3 (which is equal to the identity, so strictly speaking you didn't forget it after all ).

If x has order 9, then x^3 has order 3. Indeed, we know that x^9=e. Since (x^3)^3 = x^9=e, we see that x^3 has order 3, whereas the subgroup generated by x^3 will be of order 3. This subgroup is {e,x^3,x^6}.

If x has order 27, then x^9 has order 3. The subgroup generated by x^9 is thus {e,x^9,x^18}.