real analysis - About the differentiability of $sum_{n=1}^inftyfrac{mu(n)sin(sqrt{2n^3}x)}{sqrt{2n^3}}e^{-isqrt{n^3}x}$

Combining the identity $(1)$ from [1], I am saying the specialization $a_n=\sqrt{2n^3}$ and $b_n=i\sqrt{n^3}$ (here $i$ denotes the imaginary unit, thus $i^2=-1$) for integers $n\geq 1$, and the explanation of the criterion of Fubini's theorem, see [2] if you need it, I can prove that $$\frac{1}{\zeta(3)}=\int_0^\infty\sum_{n=1}^\infty\frac{\mu(n)\sin(\sqrt{2n^3}x)}{\sqrt{2n^3}}e^{-i\sqrt{n^3}x}dx,\tag{A}$$
where $\mu(n)$ is the Möbius function.

Question. I was wondering about questions involving this function $$f(x):=\sum_{n=1}^\infty\frac{\mu(n)\sin(\sqrt{2n^3}x)}{\sqrt{2n^3}}e^{-i\sqrt{n^3}x}$$ defined on $[0,\infty)$ that I know how solve or I don't know how solve those.

I know that the function $f(x)$ is uniformly and absolutely convergent on $[0,\infty)$, but what about the differentiability? I am asking what relevant facts we can deduce about the differentiability of our function $f(x)$ for real numbers $0\leq x<\infty$.

Many thanks.

Feel free, if you prefer, add hints for some of previous question, instead of a full answer.

References, both from this Mathematics

[1] See the answer by D'Aurizio for Cantarini's lemma, identity $(1)$ from: Find the closed form for $\int_{0}^{\infty}\cos{x}\ln\left({1+e^{-x}\over 1-e^{-x}}\right)dx=\sum_{n=0}^{\infty}{1\over n^2+(n+1)^2}$.

[2] See the second paragraph of the answer by Eldredge: When can a sum and integral be interchanged?