Assume that $f$ is multiplicative. This will ensure that we can determine $f(n)$ as the product of its values at prime powers $f(p^v).$ We set $f(1)=1.$
The values of $f(p^v)$ can be determined recursively. Start with $f(p)$, which produces the equation $ \frac{1}{p-1} = f(p) + \frac{1}{p} f(1) $ or $ f(p) = \frac{1}{p} \frac{1}{p-1}.$
Now claim that $f(p^v) = 0$ when $v\ge 2.$ Reasoning inductively, we find $ \frac{1}{p^v-p^{v-1}} = f(p^v) + \frac{1}{p^{v-1}} f(p) + \frac{1}{p^v} f(1)$ which implies $ f(p^v) = \frac{1}{p^v-p^{v-1}} - \frac{1}{p^v} \frac{1}{p-1} - \frac{1}{p^v} = \frac{p - 1 - (p-1)}{p^{v+1}-p^v} = 0.$ This shows that $f(p^v) = \begin{cases} 1 & \text{if} \quad v=0 \\ \frac{1}{p} \frac{1}{p-1} & \text{if} \quad v=1 \\ 0 & \text{otherwise.} \end{cases}$ To conclude we now identify this function. It must be zero if the square of a prime divides $n$, and positive otherwise, hence it is a multiple of $\mu^2(n).$ The denominator is simply $n\varphi(n)$, so that the end result is $ f(n) = \frac{\mu^2(n)}{n\varphi(n)}.$
The above process reflects the Dirichlet convolution $ f \star \frac{1}{n} = \frac{1}{\varphi}.$ This would suggest a possibility to compute a closed form of the function $G(s)$ from this post. However we have the Euler product $ \sum_{n\ge 1} \frac{1/\varphi(n)}{n^s} = \prod_p \left( 1 + \frac{1}{p-1} \frac{1}{p^s} + \frac{1}{p}\frac{1}{p-1} \frac{1}{p^{2s}} + \frac{1}{p^2}\frac{1}{p-1} \frac{1}{p^{3s}} + \cdots \right)$ which is $ \prod_p \left( 1 + \frac{p}{p} \frac{1}{p-1} \frac{1}{p^s} + \frac{p}{p^2}\frac{1}{p-1} \frac{1}{p^{2s}} + \frac{p}{p^3}\frac{1}{p-1} \frac{1}{p^{3s}} + \cdots \right)$ which yields in turn $\prod_p \left( 1 + \frac{p}{p-1} \frac{1/p^{s+1}}{1-1/p^{s+1}} \right)$ Now we examine the roots and the singularities of this expression as in $ 1 + \frac{p}{p-1} \frac{1/z/p}{1-1/z/p}$ getting $ z = \frac{1}{p(1-p)},$ so no closed form appears possible.