Take a positive integer $k$. Let $S \subseteq \mathbb{N}$ be the set of squarefree, and define $\displaystyle \rho_k = \mu*\omega^k$, where $ \displaystyle f* g = \sum_{d \mid n} f(d)g(n/d)$ . Show that $\rho_k(n)$ is 0 for almost all $n \in S$.
Note: $\mu(n) $ is the moebius function and $\omega(n) = |\{p \mid n, p \text{ prime} \}|$
Let $n = p_1\cdots p_r$ be the prime factorization of $n\in S$. Then, its divisors are of the form $p_{i_1}\cdots p_{i_j}$ with $\{i_1,\ldots , i_j\}\subseteq \{1,\ldots , r\}$. Notice that $\mu(p_{i_1}\cdots p_{i_j}) = (-1)^j$ and $\omega^k(p_{i_1}\cdots p_{i_k}) = j^k$. Notice that there are $\binom{r}{j}$ divisors of $n$ which have $j$ prime factors.
Thus, your $\rho_k(n)$ is $$\rho_k(n) = \displaystyle\sum_{j=0}^r \binom{r}{j}(-1)^{r-j} j^k$$
Since this is the $r$-th forward difference of $x^k$ at $x=0$, for sufficiently large $r$ it must be $0$.
However, I don't know what notion of "almost all" you're using, because there is an infinite number of values for which $\rho_k$ isn't $0$ (eg at prime numbers), and I think that "almost all" usually refers to "all but finitely many".