I am out to prove that if $$f(n)=\prod\limits_{d|n}g(d)$$ then $$g(n)=\prod\limits_{d|n}f(d)^{\mu{n\over{d}}}$$ I have been grappling with this for a bit too long now, and would really appreciate some help. I suspect logarithms might be helpful after looking at similar problems, namely this one, however I am not to familiar with their workings.
Regarding Multiplicative functions and the Möbius Inversion Formula
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You can also avoid logarithms. If $f(n)=\prod_{a\mid n}g(a)$, then
$$ \prod_{b\mid n}f(b)^{\mu(n/b)} = \prod_{b\mid n}\prod_{a\mid b} g(a)^{\mu(n/b)}=\prod_{a\mid n}\prod_{c\mid\frac{n}{a}}g(a)^{\mu\left(\frac{n}{ac}\right)} $$and for a fixed $a$ which is a divisor of $n$, $$ \sum_{c\mid \frac{n}{a}}\mu\left(\frac{n}{ac}\right)=0 $$ unless $a=n$. It follows that the previous product is exactly $g(n)$, as expected.
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The Mobius inversion formula states that for sequences $\{f_n\}_{n \geq 1}$ and $\{g_n\}_{n \geq 1}$, if $f_n = \Sigma_{d|n}g_d$, then $g_n = \Sigma_{d|n}\mu(\frac{n}{d})f_d$. There is a nice proof of this in Wilf's book Generatingfunctionology, which should be available as a free PDF file.
I think what the other logarithm based answers are saying is correct, that your version can be reduced to the form above by taking logarithms, and then applying the Mobius inversion formula, and exponentiating at the end to get your RHS. That is,
Taking logarithms (to your LHS), ${\text log\ } f_n = \Sigma_{d|n}{\text log\ } g_d$. By the inversion formula referenced above, $${\text log\ }g_n = \Sigma_{d|n}\mu(\frac{n}{d}){\log\ }f_d = \Sigma_{d|n}{\text log }f_d^{\mu(\frac{n}{d})}$$ and then exponentiating, $g_n = \Pi_{d|n}f_d^{\mu(\frac{n}{d})}$.
The comments to @posilon 's answer give some additional clarification about making sure the operations are defined: I'm just giving detail on the basic steps and don't have anything to add regarding those subtleties.
You just need to use that $\log(xy)=\log x + \log y$ and $\log(x^y)=y \log x$. Applying $\log$ to your equations (and making the necessary assumptions for the values of $g$ so that the logarithms are defined) yields the usual Möbius inversion formula.