Let $f(x)$ be defined for all rational $x$ in $0\leq x\leq 1$ $$F(n)=\sum_{k=1}^n f\bigg(\frac kn\bigg), \quad F^* (n)=\sum_{k=1\\(k,n)=1}^nf\bigg(\frac kn\bigg).$$ Prove that $$F*=\mu * F$$ where $*$ denoted the Dirichlet Multiplication and $\mu$ is the Mobius Function.
I have knowledge about Elementary Number Theory, and now I'm studying Analytic Number Theory by Apostol, which I feel very hard to self-learn, and I've seen the proof of $\phi(n)=\sum_{d\vert n} \mu(d)n/d$ which uses a very elegant way, and I think most of people cannot have that idea in mind.
By the way, I have tried when $n=6$, it shows \begin{align*}\mu* F(6)=\sum_{d\vert 6}\mu\bigg(\frac 6d\bigg)\sum_{k=1}^d f\bigg(\frac kd\bigg)\end{align*} And I just merely expanding everythings out, using the definitions of them, found it equals to $f(1/6)+f(5/6)$, it is true, but I cannot find any approach to prove it. Please help.
Thanks to @BrunoAndrades, I should first prove $F=u * F^*$, where $u(n)=1$ for all positive integer $n$, which is quite easier: \begin{align*}F(n)&=\sum_{k=1}^nf\bigg(\dfrac kn\bigg) \end{align*} Among all fractions $k/n$ where $1\leq k\leq n$, after we convert it to the simplest fraction form, it is of the form $k'/n'$ where $(k',n')=1$, then since $n'|n$ and $f\bigg(\dfrac{k'}{n'}\bigg)$ is contained in the summation of $F^*(n')$, hence we finally can conclude that $$F(n)=\sum_{d|n}\sum_{k=1\\(k,d)=1}^df\bigg(\dfrac kd\bigg)=\sum_{d|n}F^*(d)=u*F^*$$ So that Möbius Inversion Formula tell us $$F^*=\mu * F.$$