Since (.) :: (q -> r) -> (p -> q) -> (p -> r), we have f :: q -> r and swap :: p -> q.  Type unification of f requires q = (a, b) and r = c.

Since f :: (a, b) -> c and curry :: ((l, m) -> n) -> (l -> m -> n), type

unification requires l = a, b = m, and n = c.  Therefore,

curry :: ((a, b) -> c) -> (a -> b -> c), and (curry f) :: a -> b -> c.

Since flip :: (s -> t -> u) -> t -> s -> u, type unification requires

s = a, t = b, and u = c.  Therefore, flip :: (a -> b -> c) -> b -> a -> c,

and flip (curry f) :: b -> a -> c.

Therefore, curry (f . swap) ::  b -> a -> c, and p :: b -> a.  Therefore,

swap :: b -> a -> (a, b), and:

swap                       :: b -> a -> (a, b)

swap x y                   =  (y, x)