The "oscillation" of function f on interval [a, b] is sup(f)- inf(f) where where the sup and inf are for all values of x in the interval. The "oscillation" of f at a point, p, is the limit of the oscillation over intervals that include p as their length goes to 0. It should be obvious that the "oscillation" of a continuous function at a point is 0.
Given any , for x between and , 1/x takes on all values larger than and less than . That includes, of course, multiples of so sin(1/x) takes on values of 1 and -1 in that interval and the "oscillation" in that interval is 1-(-1)= 2. The limit of that, as goes to 0, is, of course, 2. That's why the oscillation of f(1/x) at x=0 is 2.
Since f(1/x) is continuous at any non-zero x, its oscillation is 0 there.