I am having some trouble trying to solve this question.
I have been learning some number theory and I have come across this question. To be honest I am kinda lost.
I have seen this being solved
Let $n \in \mathbb{Z}$ with $n > 0$. Let $F(n) = \sum_{d \mid n} \lambda(d)$. Prove that $$F(n) = \begin{cases}1, \quad \text{if }n \text{ is a perfect square}\\ 0, \quad \text{otherwise} \end{cases} $$
but I have no idea how to solve the first one.
Any help would be appreciated.
By the definition that you posted, $\lambda(n)=(-1)^{\omega(n)}$, where $\omega(n)$ is the number of prime factors of $n$, counting multiplicity. Since $\omega(mn)=\omega(m)+\omega(n)$ for all $m,n\in\mathbb N$, we have that $$\lambda(mn)=(-1)^{\omega(mn)}=(-1)^{\omega(m)+\omega(n)}=(-1)^{\omega(m)}(-1)^{\omega(n)}=\lambda(m)\lambda(n)$$ which proves that $\lambda$ is completely multiplicative. As for your summation problem, I will offer you the following hint (a list of suggested steps for solving it) and let you fill in the rest:
First prove that if $f(n)$ is a multiplicative function (not necessarily completely multiplicative), then $F(n)=\sum_{d|n}f(d)$ is also multiplicative
Then notice that if $n=\prod p_i^{m_i}$ is the prime factorization of $n$ and $f$ is a multiplicative function, then $f(n)=\prod f(p_i^{m_i})$
Finally, evaluate $F(n)$ for $n=p^k$ where $p$ is some prime, and then express a general formula for $F(n)$ for any $n$ by decomposing $n$ into its prime factorization