you should be able to solve this

G = (x) is a cyclic group, order n.
f is a divisor of n.
prove there are φ(f) elements of order f in G, where φ is the euler phi function.

In a cyclic group, order n, there is one and only one subgroup, order f, if f is a divisor of f.
(x^(n/f)) is the one you need.

Why is it the only one ? If (x^k) is another groupe of order f, then you have x^kf= e. So the order of G is a divisor of kf and then n/f is a divisor of k. But then x^k is in (x^(n/f)).
So you have (x^k) include in (x^(n/f)) and because of their equal order, (x^k)=(x^(n/f)).

So there is a unique subgroup of order f, name it H, in your fucking cyclic group (which, btw, is Z/nZ). The generators of H are all the integers a that verify : gcd(a,d) = 1 (dunno how you call it in english). They are all of order d,no one other is of order d and they are counted by the fucking euler phi function.

There you go, fgsfds