First, as you also observed, it is easier to consider the completement of the graph, so we need to count the number of 2-regular graphs on 6 vertices.
Second, we count the number of graphs isomorphic to the cycle C_6. It is 6!/(2*6)=60.
However, you missed another case: the vertices can form two disjoint triangles. There are 6!/(3!*3!*2)=10 isomorphic graphs.
One can also show that these two graphs are the only possibilities for 2-regular graphs on 6 vertices.
