How to show this equality If $f$ is  a multiplicative function and ¨$n$¨ is a square-free positive integer. Prove that:
$$f^{-1}(n) = \lambda(n)\cdot f(n)$$
where $f^{-1}$ is the dirichlet inverse and $\lambda$ is the Lioville function.
This is what i´ve try:
$$f^{-1}(n) = \frac{-1}{f(1)} \sum _{{d|n}}f\Big(\frac{n}{d}\Big)\cdot f^{-1}({d}) = - \sum _{{d|n}}f\Big(\frac{n}{d}\Big)\cdot f^{-1}({d}) $$
$$=- f(n)\sum _{{d|n}}f\Big(\frac{1}{d}\Big)\cdot f^{-1}({d})$$
 A: Let us recall  some basic facts to start.   First, with $n$ squarefree
and $n>1$, we have
$$\sum_{d|n} \lambda(d)
= \prod_{p|n} \left(1 - 1\right) = 0.$$
Second, for $n>1$ the definining relation of the Dirichlet inverse is that
$$\sum_{d|n} f(d) f^{-1}(n/d) = 0$$
so that
$$f^{-1}(n) f(1) = - \sum_{d|n, \; d>1} f(d) f^{-1}(n/d)$$
or
$$ f^{-1}(n) =
- \frac{1}{f(1)} \sum_{d|n, \; d>1} f(d) f^{-1}(n/d).$$
We now prove the claim by  induction on the number $k$ of prime factors of
$n$, which is squarefree as we point out once more. The base case is $n=p$ 
i.e. $k=1.$
We get $$f^{-1}(p)
= - \frac{1}{f(1)} f(p) f^{-1}(1) = - f(p) = \lambda(p) f(p),$$
so the claim holds.

Now  for   the  induction  step   we  suppose  $n$  has   $k+1$  prime
factors. Using the induction hypothesis we have for $d>1$ that $n/d$ has at most
$k$ prime factors so we get
$$f^{-1}(n) =
- \frac{1}{f(1)} \sum_{d|n, \; d>1} f(d) f^{-1}(n/d)
= - \frac{1}{f(1)} \sum_{d|n, \; d>1} f(d) \lambda(n/d) f(n/d).$$
Observe that since $f$ is multiplicative we have
$$f(d) f(n/d) = \prod_{p|n} f(p)$$
because with $n$ being squarefree $d$ and $n/d$ share no prime factors and all prime factors are covered, which yields
$$- \frac{1}{f(1)} \prod_{p|n} f(p)
\sum_{d|n, \; d>1} \lambda(n/d) 
= - \frac{1}{f(1)} \prod_{p|n} f(p)
\left(-\lambda(n) + \sum_{d|n} \lambda(n/d) \right).$$
Recall from the introduction that with $n$ squarefree we have
$$\sum_{d|n} \lambda(n/d) = 0$$
so this becomes
$$ - \frac{1}{f(1)} \prod_{p|n} f(p) \times - \lambda(n)
= \lambda(n) \prod_{p|n} f(p) = \lambda(n) f(n).$$
Here we have used the fact that with $f(n)$ multiplicative we have $f(1)=1.$ QED.
