Let $\displaystyle \sum\limits_{n=1}^{\infty}\frac{\sin{n^3x}}{n^2}$ be a series:
a. Find where the series converge pointwise and where uniformally.
b. Does its derivative is continuous?
About A: Using Idea from the comment the absolute value of the series is less then $\frac 1 {n^2}$ so it converges uniformly (M-test) on $\mathbb R$.
About B: I think that if the series converges and the derivative converges uniformly on $[a,b]$, it means the derivative is continuous in $[a,b]$ and it solves b. That means if I knew how to solve A then I'd be able to solve also b. Am I right?
By the definition of the derivative, letting $f(x) = \sum\limits_{n=1}^\infty \frac{\sin{n^3x}}{n^2}$, we have $$f'(x) = \lim_{h \to 0} \frac{\sum\limits_{n=1}^\infty \frac{\sin{(n^3 x)}\cos{(n^3 h)} + \cos{(n^3x)}\sin{(n^3 h)}}{n^2} - \sum\limits_{n=1}^\infty \frac{\sin{(n^3 x)}}{n^2}}{h}$$ by applying the trig identity for $\sin{(x+h)}$. By comparison to the series $\sum\limits_{n=1}^\infty \frac{C}{n^2}$ for fixed real $C$ we have that both series in the numerator converge, and by the Weierstrass M-test, in fact uniformly, so that $$f'(x) = \lim_{h\to0} \sum\limits_{n=1}^\infty \frac{\sin{(n^3x)} \cos{(n^3h)} + \cos{(n^3 x)} \sin{(n^3h)} - \sin{(n^3 x)}}{n^2 h} \; .$$
The uniform convergence of this series (again by the M-test) for $h \ne 0$ allows us to interchange the limit and sum, so that by using the formulae $$\lim_{x \to 0} \frac{\sin{x}}{x} = 1$$ and $$\lim_{x \to 0} \frac{\cos{x} - 1}{x} = 0$$ we have $$f'(x) = \sum\limits_{n=1}^\infty n \cos{n^3 x}$$ which will only converge for $\cos{n^3 x} = 0$.
The point of confusion here may be why interchanging the limit and sum is justified at the point $h = 0$. By the M-test we said that the sum was uniformly convergent on $h \in (-\infty , 0 ) \cup (0, \infty)$. Then we have that $0$ is a limit point of the set on which our sequence is uniformly convergent, and we are allowed to interchange the limit and sum (see Rudin's Principles of Mathematical Analysis, theorem 7.11 on page 149 of the third edition).