Let $\hat f(n)$ be the fourier coefficients of a a function $f$.
Let $g(x)=\int_{0}^{x}f(t)dt$.
I'm asked to find the fourier series of $g$.
Is it correct to say that since $g' (x)=f(x)$, and $\hat{g'}(n)=in\hat{g}(n)$, then $\hat{g}(n)=\frac{\hat f(n)}{in}$?
You've ignored the evaluation terms in the integration by parts: $$ 2\pi\hat{g}(0) = \int_{-\pi}^{\pi}g(t)dt=tg(t)|_{t=-\pi}^{\pi}-\int_{-\pi}^{\pi}tg'(t)dt \\ = \pi g(\pi)+\pi g(-\pi)-\int_{-\pi}^{\pi}tf(t)dt. $$ For $n \ne 0$, $$ 2\pi \hat{g}(n) = \int_{-\pi}^{\pi}g(t)e^{-int}dt=\left.g(t)\frac{e^{-int}}{-in}\right|_{t=-\pi}^{\pi}-\int_{-\pi}^{\pi}f(t)\frac{e^{-int}}{-in}dt. $$