Solution:
We claim that either $n$ is the product of distinct primes, or $n=1$.

Define $g(x)=\frac{f}{x}$. The equation we are trying to solve becomes $g(c)=\frac{1}{n}$.

The definition of the function becomes $g(p)=\frac{1}{p}$ for all primes $p$, and
$$g(a b)=g(a)+g(b)$$
Substituting $a=1$ in the above yields $g(1)=0$. Letting $b=\frac{1}{a}$ in (1) and using $g(1)=0$ gives $g(a)=-g\left(\frac{1}{a}\right)$ for all $a$ in $\mathbb{Q}^{+}$.

An easy induction then proves that
$$g\left(a^{n}\right)=n g(a)$$
for positive integers $n$ and $a$ in $\mathbb{Q}^{+}$.

This gives
$$g\left(p^{e}\right)=\frac{e}{p}$$
for prime $p$ and positive integers $e$.

These facts, combined, give us the general formula for $g\left(\frac{p_{1}^{e_{1}} \cdots p_{k}^{e_{k}}}{q_{1}^{f_{1}} \cdots q_{\ell}^{f_{\ell}}}\right)$ in terms of the prime factorizations of $p$ and $q$ :
$$g\left(\frac{p_{1}^{e_{1}} \cdots p_{k}^{e_{k}}}{q_{1}^{f_{1}} \cdots q_{\ell}^{f_{\ell}}}\right)=\frac{e_{1}}{p_{1}}+\cdots+\frac{e_{k}}{p_{k}}-\frac{f_{1}}{q_{1}}-\cdots-\frac{f_{\ell}}{q_{\ell}}=\frac{m}{p_{1} \cdots p_{k} q_{1} \cdots q_{\ell}}$$
for some integer $m$. Observe that the denominator is a product of distinct primes. Thus, if $g(c)=\frac{1}{n}$ for some $c$, then either $n$ is the product of distinct primes, or $n=1$.

It remains to prove that all such $n$ have such a solution $c$.

When $n=1$, taking $c=p^{p}$ for some prime $p$ works by (3).

We now prove that if $g(c)=\frac{1}{n}$ for some $c \in \mathbb{Q}^{+}$ and positive integer $n$, then there exists $d \in \mathbb{Q}^{+}$ such that $g(d)=\frac{1}{n p}$, for any prime $p$ relatively prime to $n$. This finishes the problem by induction.

Let $x$ and $y$ be integers whose values will be determined later. Observe that, by (2), $g\left(c^{x}\right)=x g(c)$. By (3), we get $g\left(p^{y}\right)=\frac{y}{p}$. Finally, using (1) on $c^{x}$ and $p^{y}$ gives
$$g\left(c^{x} p^{y}\right)=\frac{x}{n}+\frac{y}{p}=\frac{p x+n y}{n p}$$
It remains to choose integers $x$ and $y$ such that $p x+n y=1$. But by Bézout's identity, as the greatest common divisor of $p$ and $n$ is 1, there do exist such integers. Taking $d=c^{x} p^{y}$ then gives $g(d)=\frac{1}{n p}$, finishing the problem.