Perfect balance is not possible in the situation. There are 81 possible teams of one man and one woman, however any schedule, by definition, must have an even number of partnerships. It turns out the best solution is to play 10 rounds, and have one pair of players who never meet either as partners or opponents, and who have one less game than the others. The odd pair below is M9/F9). All the other players have 9 games each, and therefore have 9 same-sex opponents, but as there are only 8 possible same-sex opponents, one of these must be repeated (for example M1/M2 F1/F2 in the first and last rounds).
[(M3 F3):(M4 F4)] [(M7 F7):(M8 F8)] [(M5 F5):(M6 F6)] [(M1 F1):(M2 F2)]
[(M1 F8):(M4 F5)] [(M2 F6):(M7 F4)] [(M3 F2):(M8 F9)] [(M6 F7):(M9 F3)]
[(M2 F7):(M3 F6)] [(M1 F5):(M8 F3)] [(M4 F1):(M7 F9)] [(M5 F8):(M9 F4)]
[(M1 F9):(M5 F4)] [(M8 F2):(M9 F6)] [(M3 F1):(M6 F8)] [(M2 F5):(M4 F7)]
[(M2 F9):(M6 F3)] [(M7 F1):(M9 F5)] [(M4 F2):(M5 F7)] [(M1 F6):(M3 F8)]
[(M6 F4):(M7 F2)] [(M3 F7):(M5 F1)] [(M2 F3):(M9 F8)] [(M4 F9):(M8 F6)]
[(M5 F3):(M8 F1)] [(M4 F8):(M6 F2)] [(M1 F4):(M9 F7)] [(M3 F9):(M7 F5)]
[(M3 F5):(M9 F2)] [(M2 F8):(M5 F9)] [(M1 F3):(M7 F6)] [(M6 F1):(M8 F4)]
[(M4 F6):(M9 F1)] [(M1 F7):(M6 F9)] [(M2 F4):(M8 F5)] [(M5 F2):(M7 F3)]
[(M7 F8):(M8 F7)] [(M3 F4):(M4 F3)] [(M1 F2):(M2 F1)] [(M5 F6):(M6 F5)]Alternatively you could play
the spouse avoiding schedule here if you have 9 couples.
