NO

Problem:
 f(a(),f(a(),x)) -> f(f(a(),a()),a())

Proof:
 Nonconfluence Processor:
  terms: f(a(),f(f(a(),a()),a())) *<- f(a(),f(a(),f(a(),f2()))) ->* f(f(a(),a()),a())
  Qed
  
  first automaton:
   final states: {1}
   transitions:
    a() -> 7,4,3,2
    f(4,3) -> 5*
    f(5,2) -> 6*
    f(7,6) -> 1*
  
  second automaton:
   final states: {8}
   transitions:
    a() -> 11,10,9
    f(11,10) -> 12*
    f(12,9) -> 8*