Just see the special case first in $R^3$, that's $x=(x_1,x_2,x_3)$ and $u=(u_1,u_2,u_3)$, then you can handle the general case. So, we have
$$ f(u_1,u_2,u_3)= \int_{{R}^3} e^{-|x|^2}e^{-ix.u}dx$$
$$=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x_1^2+x_2^2+x_3^2)}e^{-i(x_1u_1+x_2u_2+x_3 u_3)}dx_1dx_2dx_3 $$
$$ = \int_{-\infty}^{\infty}e^{-(x_1^2+ix_1u_1)}dx_1\int_{-\infty}^{\infty}e^{-(x_2^2+ix_2u_2)}dx_2 \int_{-\infty}^{\infty}e^{-(x_3^2+ix_3u_3)}dx_3 $$
$$ = \prod_{k=1}^{3}\int_{-\infty}^{\infty}e^{-(x_k^2+ix_ku_k)}dx_k = \prod_{k=1}^{3}\sqrt{\pi}e^{-\frac{1}{4}u_k^2 } = {\pi}^{\frac{3}{2}}e^{-\frac{1}{4}(u_1^2+u_2^2+u_3^2)}= {\pi}^{\frac{3}{2}}e^{-\frac{1}{4}|u|^2}.$$
Now, you can figure out the general case easily. Note that, for evaluating the above integrals, we first competed the square then use the Gaussian integral.