YES
We show the termination of the TRS R:

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


-- SCC decomposition.

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

p1: f#(s(x),y,y) -> f#(y,x,s(x))

and R consists of:

r1: g(x,y) -> x
r2: g(x,y) -> y
r3: f(s(x),y,y) -> f(y,x,s(x))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: f#(s(x),y,y) -> f#(y,x,s(x))

and R consists of:

r1: g(x,y) -> x
r2: g(x,y) -> y
r3: f(s(x),y,y) -> f(y,x,s(x))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          f#_A(x1,x2,x3) = 1
          s_A(x1) = x1 + 2
          g_A(x1,x2) = x1 + x2 + 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          f#_A(x1,x2,x3) = max{0, x1 - 8, x2 - 6, x3 - 8}
          s_A(x1) = max{7, x1 + 2}
          g_A(x1,x2) = max{x1, x2}
          f#!_A(x1,x2,x3) = max{3, x1 - 3, x2 - 2, x3 - 5}
          s!_A(x1) = max{1, x1 - 7}
          g!_A(x1,x2) = max{x1, x2}
  
    partial status:
  
      pi(f#) = []
      pi(s) = [1]
      pi(g) = [1, 2]

The following rules are removed:

  p1

Then no dependency pair remains.