Suppose all the positive divisors of $n$ are $d_1 < d_2 < \dots < d_k$. Then
$$f(n) = \frac{d_1 + d_2 + \dots + d_k}{n} = \frac{d_1}{n} + \frac{d_2}{n} + \dots + \frac{d_k}{n} = \frac{1}{d_k} + \frac{1}{d_{k-1}} + \dots + \frac{1}{d_1}.$$
Thus $f(n)$ is the sum of the reciprocals of the positive divisors of $n$. Because $768 = 384 \cdot 2$, all the divisors of $384$ also divide $768$, so it is necessary to consider only the divisors of $768$ that do not divide $384$, which are $2^8$ and $2^8 \cdot 3$. Therefore
$$f(768) - f(384) = \frac{1}{2^8} + \frac{1}{2^8 \cdot 3} = \frac{4}{2^8 \cdot 3} = \frac{1}{192}.$$

Let $\sigma(n)$ be the sum of the divisors of $n$. Then $f(n) = \frac{\sigma(n)}{n}$, and because $\sigma$ is multiplicative (meaning that $\sigma(ab) = \sigma(a)\sigma(b)$ whenever $\gcd(a, b) = 1$), $f$ is multiplicative as well. That is,
$$f(768) - f(384) = f(2^8 \cdot 3) - f(2^7 \cdot 3) = f(3)(f(2^8) - f(2^7)).$$
Note that $f(3) = \frac{4}{3}$ and
$$f(2^k) = \frac{\sigma(2^k)}{2^k} = \frac{2^{k+1} - 1}{2^k} = 2 - \frac{1}{2^k}.$$
Thus
$$f(3)(f(2^8) - f(2^7)) = \frac{4}{3} \cdot \frac{1}{2^8} = \frac{1}{2^6 \cdot 3} = \frac{1}{192}.$$