Let $k(n)=\prod_{p \mid n}p$ be the kernel of the integer, so that $f(n)=n/k(n)$. As indicated in p. 7 of Finch's article,
$$\sum_{n\le x} \frac{1}{k(n)} = \exp\left( \left( \frac{8\log x}{\log \log x}\right)^{1/2}(1+o(1))\right)$$
was shown by de Bruijn in "On the number of integers $\le x$ whose prime factors divide $n$", Illinois J. Math. 6 (1962) 137–141. This solved a question of Erdős.
The proof went by first showing that $F(s)=\sum_n \frac{1}{k(n)} n^{-s}$ behaves likes $s^{-1} (\log (s^{-1}))^{-1}$ as $s \to 0^+$, and then applying a Tauberian theorem.

De Bruijn and van Lint showed that
$$\sum_{n \le x} \frac{n}{k(n)} = o\left( x \sum_{n \le x} \frac{1}{k(n)}\right)$$
in "On the number of integers $\le x$ whose prime factors divide $n$", Acta Arith. 8 (1963) 349–356. See their paper for the motivation. Combining with the result of de Bruijn, this gives an upper bound for $\sum_{n \le x} n/k(n)$ agreeing with Will Sawin's heuristic result from the comments.

You might also be interested in their paper "On the asymptotic behaviour of some Dirichlet series with a complicated singularity" (Nieuw Archief voor Wiskunde, 3/11, 68-75, 1963).

Wolfgang K. Schwarz in "Einige Anwendungen Tauberscher Sätze in der Zahlentheorie. B" (J. Reine Angew. Math. 219, 157-179 (1965)) determined the asymptotics of $\sum_{n \le x} \frac{1}{k(n)}$ up to $(1+o(1))$ in terms of an implicit quantity (this is in the first page of his paper, but see Satz III.17 for a more explicit result that is less accurate). He also recovers the $o(1)$ result of de Bruijn and van Lint (Satz III.16).

There might be an earlier reference for a lower bound for $\sum_{n \le x} n/k(n)$, but the one I can find follows from the the 113-page paper "Sur la répartition du noyau d’un entier" by Robert and Tenenbaum (Indag. Math., New Ser. 24, No. 4, 802-914 (2013)). They denote $\sum_{n \le x} n/k(n)$ by $K_2(n)$ and estimate it in Théorème 4.4, which gives
$$\sum_{n \le x} \frac{n}{k(n)} \sim x \sqrt{\frac{2}{\log x \log \log x}}\sum_{n \le x} \frac{1}{k(n)}.$$
Combining this with Schwarz's work gives an asymptotic formula for your sum.