The sum of six can be achieved in the following ways: (1, 5), (2, 4), (3, 3), (4, 2), and (5, 1). Jane can roll doubles in the following ways: (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6). There is only one way to get a sum of six with Joe and only one way to roll doubles with Jane. So, the probability that Joe rolls a sum of six and Jane rolls doubles is $\dfrac{1\cdot1}{6\cdot6} = \boxed{\dfrac{1}{36}}$.
