You only have to prove the existence of a non-trivial normal subgroup in a group , since then both this sbgp. and the resulting factor group are solvable and thus also is (assuming you already know/can prove that any group of order less than 36 is solvable).
Assume there are 4 Sylow 3-sbgps.: if all of them have trivial intersection we have elements of order 3 or 9, and then the other 4 elements must be the unique Sylow 2-sbgp. which is then normal.
So assume two different Sylow 3-sbgps. It then must be that the normalizer of has at least 15 elements (all the ones in both ...why?) , so it at least has order 18 (why?), and thus either this normalizer sbgp. or are normal non-trivial sbgps. of and we're done.
Ps. It can be proved, but it's lenghty, that any group of order 36 has at least one Sylow sbgp. normal.