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.