Here candidate keys are AB ,BC and F.
There is partial dependency in FD ( B->D ).
I think C->A is not partial dependency ,since A is key attribute.
So to eliminate that divide the table into two tables as
take B+ = { B,D,E } as one table and remaining as others and B is the common attribute .
So (ABCF) , (BDE) .
Correct if anything wrong.