YES

We show the termination of R/S, where R is

  f(a,g(y),z) -> f(a,y,g(y))
  f(b,g(y),z) -> f(a,y,z)
  a -> b

and S is:

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

Since R almost dominates S and S is non-duplicating,
it is enough to show finiteness of (P, Q).
Here P consists of the dependency pairs

  f#(a,g(y),z) -> f#(a,y,g(y))
  f#(a,g(y),z) -> a#
  f#(b,g(y),z) -> f#(a,y,z)
  f#(b,g(y),z) -> a#

and Q consists of the rules:

  f(a,g(y),z) -> f(a,y,g(y))
  f(b,g(y),z) -> f(a,y,z)
  a -> b
  f(x,y,z) -> f(x,y,g(z))
  f#(x,y,z) -> f#(x,y,g(z))

The weakly monotone algebra (N^2, >_lex) with

  f#_A((x1,y1),(x2,y2),(x3,y3)) = (x2, 1)
  a_A = (2, 1)
  g_A((x1,y1)) = (x1 + 1, y1 + 1)
  a#_A = (0, 0)
  b_A = (1, 2)
  f_A((x1,y1),(x2,y2),(x3,y3)) = (x2, 0)

strictly orients the following dependency pairs:

  f#(a,g(y),z) -> f#(a,y,g(y))
  f#(a,g(y),z) -> a#
  f#(b,g(y),z) -> f#(a,y,z)
  f#(b,g(y),z) -> a#

We remove them from the set of dependency pairs.

No dependency pair remains.