NO

Problem:
 f(f(x)) -> f(g(f(f(x))))
 f(x) -> x

Proof:
 Nonconfluence Processor:
  terms: f(g(f(f(f3())))) *<- f(f(f3())) ->* f(f3())
  Qed
  
  first automaton:
   final states: {8}
   transitions:
    f3() -> 9*
    f(10) -> 11*
    f(9) -> 10*
    f(12) -> 11,8
    g(11) -> 12*
    12 -> 11,8
    9 -> 10*
    10 -> 11*
  
  second automaton:
   final states: {13}
   transitions:
    f3() -> 14*
    f(14) -> 13*
    14 -> 13*