What is the difference between (fo(goh)(x) and ((fog)oh)(x)?

is associative that is
##(f@(g@h))(x) = ((f@g)@h)(x)##
There is no difference in the result though the steps may be expressed differently.
##(f@(g@h))(x) = f(g(h(x))) = ((f@g)@h)(x)##
For example suppose:
##f(x) = x^2##
##g(x) = 1/x##
##h(x) = x + 1##
Then:
##(g@h)(x) = g(h(x)) = 1/(x+1)##
##(f@(g@h))(x) = f((g@h)(x)) = f(1/(x+1)) = 1/(x+1)^2##
##(f@g)(x) = f(g(x)) = 1/x^2##
##((f@g)@h)(x) = (f@g)(h(x)) = (f@g)(x+1) = 1/(x+1)^2##