Actually, for a simple case (when f(x) is a continues function and xf(x) is absolutly integrable on the real line), you can just insert the differentiation inside the integral, that is $ \frac{d}{dy} \int_{-\infty}^{\infty} f(x)e^{-ixy}dx = \int_{-\infty}^{\infty} \frac{d}{dy} (f(x)e^{-ixy})dx = \int_{-\infty}^{\infty} -ixf(x)e^{-ixy} dx = -i \int_{-\infty}^{\infty} xf(x)e^{-ixy} dx $ since it can be shown that both $ f(x)e^{-ixy} $ and $ -ixf(x)e^{-ixy} $ are continues, and the integral $ \int_{-\infty}^{\infty} -ixf(x)e^{-ixy} dx $ converges uniformly (for every y) .
for functions that are not continues, other techniques are required, like Lebesgue's dominated convergence theorem.