NO

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

Proof:
 Nonconfluence Processor:
  terms: f(g(g(f2()))) *<- f(g(f(g(f(f2()))))) ->* g(g(f(f2())))
  Qed
  
  first automaton:
   final states: {1}
   transitions:
    f(4) -> 1*
    g(3) -> 4*
    g(2) -> 3*
    f2() -> 2*
  
  second automaton:
   final states: {5}
   transitions:
    f(6) -> 7*
    g(8) -> 5*
    g(7) -> 8*
    f2() -> 6*