Differentiation of $\sum_{n=0}^{\infty}\frac{1}{(n+1)3^n}x^{n+1}$

Yes, that can be one way. You will get a geometric progression:

$$f'(x) = 1+ \frac{x}{3} + \frac{x^2}{9} + \frac{x^3}{27} ...$$

So $f'(x) = \frac{1}{1-x/3}$, or $f'(2) = 3$.

$$\begin{align} f(x)&=\sum_{n=0}^\infty\frac 1{(n+1)3^n}x^{n+1}\\ f'(x)&=\frac d{dx}\sum_{n=0}^\infty\frac 1{(n+1)3^n}x^{n+1}\\ &=\sum_{n=0}^\infty\frac d{dx}\frac 1{(n+1)3^n}x^{n+1} &&\text{(by Fubini/Tonelli's theorem)}\\ &=\sum_{n=0}^\infty\frac 1{(n+1)3^n}\cdot (n+1)x^n\\ &=\sum_{n=0}^\infty \left(\frac x3\right)^n\\ &=\frac 1{1-\frac x3}\\ \therefore f'(2)&=3 \end{align}$$