YES

We show the termination of the TRS R:

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

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: g#(f(x),y) -> h#(x,y)
p2: h#(x,y) -> g#(x,f(y))

and R consists of:

r1: g(f(x),y) -> f(h(x,y))
r2: h(x,y) -> g(x,f(y))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: g#(f(x),y) -> h#(x,y)
p2: h#(x,y) -> g#(x,f(y))

and R consists of:

r1: g(f(x),y) -> f(h(x,y))
r2: h(x,y) -> g(x,f(y))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      g#_A(x1,x2) = x1 + ((0,0,0),(0,0,0),(1,0,0)) x2
      f_A(x1) = ((1,0,0),(0,1,0),(1,1,1)) x1 + (2,0,0)
      h#_A(x1,x2) = ((1,0,0),(0,0,0),(1,0,0)) x1 + (1,1,3)

The next rules are strictly ordered:

  p1, p2

We remove them from the problem.  Then no dependency pair remains.