Let for integers $k\geq 1$, the Möbius function denoted by $\mu(k)$, and $M(n)=\sum_{k\leq n}\mu(k)$ the Mertens function, then one can prove easily that
$$\sum_{k=1}^n\mu(k)\frac{e^{\mu(k)}+1}{e^{\mu(k)}-1}=M(n)+2\sum_{k=1}^n\frac{\mu(k)}{e^{\mu(k)}-1}.$$
I've used the first identity in the proof of Proposition 6.9.1, page 130 of George Boros and Victor Moll, Irresistible integrals. Symbolic, Analysis and Experiments in the Evaluation of Integrals, Cambridge (2004).
Fact. The graphs of the arithmetic functions $f(n)=\sum_{k=1}^n\mu(k)\frac{e^{\mu(k)}+1}{e^{\mu(k)}-1}$ and $g(n)=2\sum_{k=1}^n\frac{\mu(k)}{e^{\mu(k)}-1}$ have slopes very well defined.
My
Question. It is possible obtain the asymptotic behaviour (you can work with the asymptotic equivalence $\sim$ or some oh notation) for some of previous arithmetica functions? I say or well for $f(n)$ or for $g(n)$ as $n\to\infty$ (truly the graphs of these functions seems the same). THanks in advance.
assuming you meant $\mu(k) \frac{e^{\mu(k)}+1}{e^{\mu(k)}-1} = 0$ when $\mu(k) = 0$ : $$\sum_{k=1}^n \mu(k) \frac{e^{\mu(k)}+1}{e^{\mu(k)}-1} = \frac{e+1}{e-1} \sum_{k\le n, \mu(k) = 1} 1 -\frac{e^{-1}+1}{e^{-1}-1} \sum_{k\le n, \mu(k) = -1} 1 $$ $$= \frac{e+1}{e-1} \sum_{k\le n, \mu(k) = 1} 1 -\frac{1+e}{1-e} \sum_{k\le n, \mu(k) = -1} 1 = \frac{e+1}{e-1} \sum_{k=1}^n |\mu(k)| \sim\frac{e+1}{e-1} \frac{6}{\pi^2} n $$
the last asymptotics being because $$\frac{\zeta(s)}{\zeta(2s)} = \prod_p \frac{1-p^{-2s}}{1-p^{-s}} = \prod_p 1+p^{-s} = \sum_{n=1}^\infty |\mu(n)| n^{-s} = s \int_1^\infty \sum_{n \le x} |\mu(n)|x^{-s-1}dx$$ is a Dirichlet series with non-negative coefficients, converging absolutely for $Re(s) > 1$ and with a pole of order $1$ and residue $\frac{1}{\zeta(2)} = \frac{6}{\pi^2}$ at $s=1$