Show that $\dfrac{\phi (n)}{n}=\sum\limits_{d|n} \dfrac{\mu (d)}{d}$ This problem is from Ram Murtys book Problems in analytic number theory. In his solutions it only says that this is an immediate result when combining the Möbius inversion formula: $$f(n)=\sum\limits_{d|n}g(d) \ \ \ \forall n\in \mathbb{N} \Leftrightarrow g(n)=\sum\limits_{d|n}\mu (d)f(n/d) \ \ \ \forall n\in \mathbb{N}$$ and gauss theorem: $$\sum\limits_{d|n}\phi (d)=n.$$ I cant get it right though...
Here is one thing that I have tried thus far:
Let $f(n)=\phi (n)$ in Möbius inversion formula. We wish to show that $g(d)\stackrel{?}{=}\dfrac{\mu (d)}{d}$. We have that $$g(n)=\sum\limits_{d|n}\mu (d)f(n/d)=\sum\limits_{d|n}\mu (d)\phi (n/d).$$ From here I have tried a bunch of things which seemed unjustified and did not actually get me anywhere.
Can someone give me a hint on how to solve this problem?

$$ {\rm Id}(n)=n=\sum_{d|n}\varphi(d)=(\varphi*1)(n) $$
Therefore using Möbius inversion formula, we have
$$ \varphi(n)=({\rm Id}*\mu)(n)=\sum_{d|n}\mu(d)\frac{n}{d}=n\sum_{d|n}\frac{\mu(d)}{d} $$
