Solution:
We claim the series reduces to $\sigma(n)$. The series counts the ordered triples $(d, x, y)$ with $d \mid n;\ x \mid d;\ 0 < y \leq n/d;$ and $(y, n/d) = 1$. To see this, write
$$\sum_{d \mid n} \phi(d) \tau\left(\frac{n}{d}\right) = \sum_{d' \mid n} \phi\left(\frac{n}{d'}\right) \tau\left(d'\right)$$
so that for a given $d' \mid n$ we may choose $x$ and $y$ as described above. On the other hand, we can count these triples by groups sharing a given $x$. Fixing $x$ as a divisor of $n$ fixes an integer $\frac{n}{x}$. Then $d$ varies such that $\frac{n}{d}$ is a divisor of $\frac{n}{x}$. For each divisor $\frac{n}{d}$ of $\frac{n}{x}$ there are precisely $\phi\left(\frac{n}{d}\right)$ choices $y$, so that by the lemma from the previous problem, there are $\frac{n}{x}$ triples $(d, x, y)$ for a given $x$. It follows that there are precisely $\sigma(n)$ such triples $(d, x, y)$.

Again, an alternative is to use the multiplicativity of the convolution, although it is now a little more difficult. Write $n = p^{k}$ so that
$$\begin{aligned} \sum_{d \mid n} & \phi(d) \tau\left(\frac{n}{d}\right) = \sum_{m=0}^{k} \phi\left(p^{m}\right) \tau\left(p^{k-m}\right) = k+1 + \sum_{m=1}^{k} p^{m-1}(p-1)(k-m+1) \\ & = k+1 + \left(\sum_{m=1}^{k} p^{m}(k-m+1)\right) - \left(\sum_{m=1}^{k} p^{m-1}(k-m+1)\right) = k+1 + p^{k} - k + \sum_{m'=1}^{k-1} p^{m'} \\ & = 1 + p + \cdots + p^{k} = \sigma\left(p^{k}\right) \end{aligned}$$