How can one show that $$\sum_{n \leq X} \frac{\mu(n)^2 \tau_k(n)}{\phi(n)} \ll (\log X)^k \ ?$$ Here, $$\tau_k(n) = \sum_{\substack{d_1,d_2, \dots, d_k \in \mathbb{N} \\ d_1d_2\cdots d_k=n}} 1.$$ I'm aware of the proof (by induction on $k$) that $\tau_k(n) \ll X(\log X)^{k-1}.$
If $n$ is squarefree, we have by multiplicativity $$\frac{\tau_k(n)}{\phi(n)} = \prod_{p \mid n} \frac{k}{p-1}.$$
We have $$\begin{align*}\sum_{n \text{ squarefree }\leq x}\prod_{p\mid n}\frac k{p-1} & \leq \prod_{p\leq x}\left(1+\frac k{p-1}\right) \\ & = \prod_{p\leq x}\left(1+\frac1p\right)^k \cdot \left(1+O(1/p^2)\right) \end{align*}$$ The product $\prod_{p\leq x}\left(1+O(1/p^2)\right)$ is bounded (even has a limit).
The product $\prod_{p\leq x}\left(1+1/p\right)$ is $\ll \log x$ by a theorem of Mertens.
For more detail, Mertens' theorem gives $$\prod_{p\leq x} \left( 1-\frac1p \right) = \frac{e^{-\gamma}}{\log x} \left( 1+O(1/\log x) \right) $$
Divide by $\prod_{p\leq x} ( 1-1/{p^2} ) = 1/\zeta(2)+O(1/x)$ to get $$\prod_{p\leq x} \left( 1+\frac1p \right) = e^{\gamma}/\zeta(2)\log x +O(1) $$