On calculations for $\prod_{k=2}^\infty \left( 1+\frac{\mu(k)}{k^3} x\right)$, where $\mu(n)$ is the Möbius function I define for some set of real numbers $x\in S$ (see that it is my Question 1.) the domain of the function $$f(x)=\prod_{k=2}^\infty \left( 1+\frac{\mu(k)}{k^3} x\right) ,$$
where $\mu(k)$ is the Möbius function.

Question 1. I am interesting in this function since I believe that it is possible justify $f'(0)=\frac{1}{\zeta(3)}-1$. If it is well known in the literature or you know how compute its domain $S$, it is for which $x$ is defined, can you tell me?

My calculations were that $0<\frac{7}{8}\leq 1+\frac{\mu(k)}{k^3}<1$ for integers $k\geq 2$, and I am stuck with what theorem relating infinite products and series can I use to answer previous question.
Subsequents calculations provide to me, 
informally $$\log f(1)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\sum_{k=2}^\infty\frac{(\mu(k))^n}{k^{3n}}.$$
When I've splitted the series for odd and even $n's$, I can write $\log f(1)$ as the sum  two terms, the first $$\sum_{\text{n odd}}\frac{(-1)^{n+1}}{n}\sum_{k=2}^\infty\frac{(\mu(k))^n}{k^{3n}}=\sum_{j=1}^\infty\frac{\zeta(3(2j-1)-1)}{2j-1},$$ and the second as $$\sum_{\text{n even}}\frac{(-1)^{n+1}}{n}\sum_{k=2}^\infty\frac{(\mu(k))^n}{k^{3n}}=-\sum_{j=1}^\infty\frac{1}{2j}\sum_{k=2}^\infty\frac{ \left| \mu(k) \right| }{k^{6j}}.$$

Question 2. Were justified my calculations for $\log f(1)$? Only is requiered a yes or where was my mistake. Also if my calculations were right if you know how write the series inside $$-\sum_{j=1}^\infty\frac{1}{2j}\sum_{k=2}^\infty\frac{ \left| \mu(k) \right| }{k^{6j}}$$
  with zeta values, as I've computed for the case $\text{ n odd}$, then these nice computations will be here as reference for all us. Thanks in advance.

 A: Answer to Question 1
First, clearly for every $x$ there exists some $N\in \mathbb{N},$ such that $\frac{|x|}{k^3}<1$ for every $k>N.$ 
Since both $\sum_{k=2}^{\infty}\frac{\mu(k)}{k^3}x$ and $\sum_{k=2}^{\infty}\frac{|\mu(k)|^2 x^2}{k^6}$ converge, using the convergence criterion of infinite products, we can see that the infinite product definition of $f(x)$ converges for all $x\in\mathbb{R}.$
Thus, domain of $f(x)$ is the entire $\mathbb{R}.$
For detailed account of the convergent criterion, see criterion. 
To calculate the derivative of a function $f$, we can take the derivative of its logarithm, 
$$
(\log f)^\prime=\frac{f^\prime}{f}.
$$
Let 
$$
P_n=\prod_{k=2}^{n}\left
(1+\frac{\mu(k)}{k^3}x\right).
$$
We have
\begin{align*}
\log f(x)&=\log\lim\limits_{n\to\infty}P_n\\&=\lim\limits_{n\to\infty}\log P_n.
\end{align*}
The first line is due to the definition of $P_n$ while the second line is due to the continuity of $\log x$.
Using Weierstrass M-test, it can be shown that both $\log P_n$ and $\frac{d}{dx}\log P_n$ converge uniformly, 
so we can take the derivative term by term,
$$
(\log f(x))^\prime=\sum_{k=2}^{\infty}\frac{d}{dx}\log\left(1+\frac{\mu(k)}{k^3}x\right)=\sum_{k=2}^{\infty}\frac{\frac{\mu(k)}{k^3}}{1+\frac{\mu(k)}{k^3}x}.
$$
In conclusion, we have
\begin{align*}
f^\prime(0)=f(0)(\log f)^\prime(0)=\sum_{k=2}^{\infty}\frac{\mu(k)}{k^3}=\frac{1}{\zeta(3)}-1.
\end{align*}
For the Weierstrass M-test part, clearly for every $x$ there exists some $L\in \mathbb{N},$ such that $\frac{|x|}{k^3}<\frac{1}{2}$ for every $k>L.$ 
For $|x|<\frac{1}{2}$, we have
\begin{align*}
|\log(1+x)|&\le |x|+|x|\bigl|\sum_{n\ge 2}\frac{(-1)^{n+1}x^{n-1}}{n}\bigr|\\
&\le |x|+|x||\sum_{n\ge 1}2^{-n}|\\
&\le 2|x|,
\end{align*}
and 
\begin{align*}
|\frac{1}{1+x}|<\frac{1}{1-|x|}<2.
\end{align*}
For $k>L,$ we have
$$\biggl|\log\left
(1+\frac{\mu(k)}{k^3}x\right)\biggr|<\frac{2|x|}{k^3},$$
and $$\biggl|\frac{d}{dx}\log\left
(1+\frac{\mu(k)}{k^3}x\right)\biggr|<\frac{2}{k^3}.$$
Using these two estimates as the respective $M_k$ for $k>L$ and considering that adding $L$ finite terms doesn't change the convergence property of a series, we can see that both $\log P_n$ and $\frac{d}{dx}\log P_n$ converge uniformly for any finite interval.
A: Since $$\sum_{k\geq2}\left|\frac{\mu\left(k\right)x}{k^{3}}\right|<\infty
 $$ for all $x
 $ the product converges for all $x
 $. If $n
 $ is even we have $$ -\sum_{n\geq1}\frac{1}{2n}\sum_{k\geq2}\frac{\left|\mu\left(k\right)\right|}{k^{6n}}=-\sum_{n\geq1}\frac{1}{2n}\left(\frac{\zeta\left(6n\right)}{\zeta\left(12n\right)}-1\right)
 $$ since $$\sum_{k\geq1}\frac{\left|\mu\left(k\right)\right|}{k^{s}}=\frac{\zeta\left(s\right)}{\zeta\left(2s\right)},\,\textrm{Re}\left(s\right)>1
 $$ and if $n
 $ is odd we have $$\sum_{n\geq1}\frac{1}{2n-1}\sum_{k\geq2}\frac{\mu\left(k\right)}{k^{6n-3}}=\sum_{n\geq1}\frac{1}{2n-1}\left(\frac{1}{\zeta\left(6n-3\right)}-1\right)
 $$ since $$\sum_{k\geq1}\frac{\mu\left(k\right)}{k^{s}}=\frac{1}{\zeta\left(s\right)},\,\textrm{Re}\left(s\right)>1.
 $$
