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For $n$ a positive integer, let $f(n)$ be the squarefree part of $n$.

Find the Euler product for $\mathfrak D_{f}(s)$ where $\mathfrak D_{f}(s)$ is the Dirichlet Series of $f$.

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Are you familiar with the general Euler product for a multiplicative arithmetic function?

$$D_{f}(s)=\prod_{p}\sum_{k\geq 0}{\frac{f(p^{k})}{p^{ks}}}$$

Throughout this, I'm going to assume convergence of the various sums - I'll leave you to work out where everything is defined. Then, for a prime $p$, the squarefree part of $p^{k}$ is $f(p^{k})=1$ if $k$ is even, and $p$ if $k$ is odd. So, putting this into the formula, we get

$$D_{f}(s)=\prod_{p}\sum_{k\geq 0}{(\frac{1}{p^{2ks}}+\frac{p}{p^{(2k+1)s}}})=\prod_{p}(\frac{1}{1-p^{2s}}+\frac{p/p^{s}}{1-p^{2s}})=\prod_{p}\frac{p^{s}+p}{p^{s}(1-p^{2s})}.$$

Here, I've used the sum of a geometric series, which you'll need to justify the convergence of.

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