Apparently, by Weierstrass' test, this function is continuous. How to prove that it is not differentiable at $x=0$? I plotted its graph. It seems that
$$\lim_{x\rightarrow 0} \frac{f}{x} =\infty . $$
How to prove it?
Term-by-term differentiation is not necessarily valid here.
One can show (by using Abel's summation and the Taylor series of the main branch of the logarithm) that $$\sum_{n=1}^\infty \frac{\cos(nx)}{n} = -\log(2\sin(x/2))$$ for $ 0 < x < \pi$ and the convergence is uniform on any interval $[a,b]$ with $0<a<b<\pi$. Thus we find for $x>0$ that (by interchanging integration and summation) $$\frac{f(x+h)-f(x)}{h} = \frac{1}{h} \int_x^{x+h} -\ln(2 \sin(t/2)) dx.$$ One can check that the integrand on the right-hand side is absolute integrable, say on $[0,\pi]$. Thus the the integral as a function of $x$ is contionous in $x=0$. We get $$\frac{f(h)-f(0)}{h} = \frac{1}{h} \int_0^{h} - \ln(2 \sin(t/2)) \, dt.$$ Since $t/2 \le 2 \sin(t/2) \le t$ for $h \in [0,\pi/2]$ and $x \mapsto \log(x)$ is monotone on $(0,1]$, we can compare the growth of the last integral with $$\frac{1}{h} \int_0^{h} \ln(t) \, dt = \ln(h)-1.$$ Thus $$\frac{f(h)-f(0)}{h} \sim -\ln(h)$$ for $h \downarrow 0$.