arithmetic functions with $ 1/ \phi(n)$ i have to find a function $f(n)$ such that
$$1/\phi(n)= \sum_{d\mid n} \frac{1}{d} f(\frac{n}{d}). $$
However, we know that :
 $$  \phi(n)= \mu* \operatorname{Id}(n) $$
$$ \phi(n)= \sum_{d\mid n}\frac{n}{d}\mu(d) $$
I have a feeling i need to use these equations but not sure how. 
 A: Define $g(n)=nf(n)$ then:
$$n/\phi(n)=\sum_{d\mid n} g\left(\frac nd\right)=\sum_{d\mid n} g(d)$$
What is $g(n)$?
A: Completing Thomas Andrews's answer.
$$
\frac{n}{\phi(n)} = \sum_{d|n}\frac{n}{d}f\left(\frac n d \right)
=\sum_{d|n}d \, f(d). \qquad (1)
$$
By looking up the inversion table on this page (the 9th equation from the top of the section):
$$
d \, f(d) = \frac{ \mu^2(d) }{ \phi(d) }.
$$
The proof can be found on pages 8-10 in Dineva.
So
$$
f(n) = \frac{ \mu^2(n) }{ n \, \phi(n) }.
$$

Edit. No cheating solution under Thomas Andrews's instruction.
Inverting (1), we get
$$
n f(n) = \sum_{d|n} \frac{d}{\phi(d)} \mu\left( \frac{n}{d} \right).
$$
This function is multiplicative since it is the Dirichlet convolution of two multiplicative functions.  Thus we only have to compute its value of a prime power
$$
\begin{aligned}
1 f(1) &= 1, \\
p f(p) &= 1 \cdot (-1) + \frac{1}{1-1/p} 1 = \frac{1}{p-1} = \frac{1}{\phi(p)}, \\
p^e f(p^e) &= \frac{1}{1-1/p} \cdot (-1) + \frac{1}{1-1/p} 1 = 0. \qquad (e \ge 2)
\end{aligned}
$$
So
$$
p^e f(p^e) = \frac{ \mu^2(p^e) } { \phi(p^e) }.
$$
Multiplying all prime factors, we get
$$
n f(n) = \frac{ \mu^2(n) } { \phi(n) }.
$$

Thanks again to J.M. for the interesting question, and Thomas Andrews's patience.
Happy thanksgiving. :-)
