Let $f:\mathbb{R}\to\mathbb{C}$, $2\pi$ periodic function and $f\in C^1$, such that the n-th Fourier coefficient is: $\hat{f}(n) = 3^{-n^2}$.
Find the Fourier coefficients of $g(x) = \pi f'(x+2009)$.
Since $f$ is $C^1$, we know that: $$ f = \sum_{n=-\infty}^\infty \hat{f}(n)e^{int} + \hat{f}(0)e^0 = 1 + \sum_{n=-\infty}^\infty 3^{-n^2}e^{int} = 1 + \sum_{n\ge 1} \frac{e^{int} + e^{-int}}{3^{n^2}} = 1 + \sum_{n\ge 1} \frac{\cos (nt)}{2\cdot 3^{n^2}}$$
It's easy to see using M-test that $f$ converges uniformly and absolutely. Hence,
$$f' = -\frac{1}{2} \sum_{n\ge 1} \frac{\sin(nt)}{3^{n^2}}$$
Applying our result for $g(x)$ we have that $$g(x) = \frac{\pi}{2}\sum_{n\ge 1} \frac{\sin(nx + 2009x)}{3^{n^2}} = \frac{\pi}{2} \sum_{n\ge 1} \frac{e^{-inx -2009ix} -e^{inx + 2009ix}}{2\cdot 3^{n^2}} = \frac{\pi}{4} \sum_{n=-\infty}^\infty \frac{e^{i(n+2009)x}}{3^{n^2}}$$
So basically I have $g(x)$ as an infinite sum of exponents.
It looks like $\hat{g}(n) = \hat{f}(n-2009)$. Am I right?
I don't think so - it looks like you are missing a factor of $n$. I think it's a little simpler than this if you stick with exponentials. Write
$$f(x) = \sum_{n=-\infty}^{\infty} 3^{-n^2} e^{i n x} $$
This converges, as you said, absolutely and uniformly. We may write
$$f'(x) = i \sum_{n=-\infty}^{\infty} n \, 3^{-n^2} e^{i n x} $$
$$g(x) = \pi f'(x+2009) = i \pi \sum_{n=-\infty}^{\infty} n \, 3^{-n^2} e^{i 2009 n} e^{i n x} $$
You should be able to read the Fourier coefficients right away from this.