YES

Problem:
 f(x,y) -> g(x,y)
 g(h(x),y) -> h(f(x,y))
 g(h(x),y) -> h(g(x,y))

Proof:
 DP Processor:
  DPs:
   f#(x,y) -> g#(x,y)
   g#(h(x),y) -> f#(x,y)
   g#(h(x),y) -> g#(x,y)
  TRS:
   f(x,y) -> g(x,y)
   g(h(x),y) -> h(f(x,y))
   g(h(x),y) -> h(g(x,y))
  Usable Rule Processor:
   DPs:
    f#(x,y) -> g#(x,y)
    g#(h(x),y) -> f#(x,y)
    g#(h(x),y) -> g#(x,y)
   TRS:
    
   KBO Processor:
    argument filtering:
     pi(h) = [0]
     pi(f#) = [0]
     pi(g#) = 0
    usable rules:
     
    weight function:
     w0 = 1
     w(f#) = w(h) = 1
     w(g#) = 0
    precedence:
     h > g# ~ f#
    problem:
     DPs:
      
     TRS:
      
    Qed