I have to calculate Fourier transform of $f:\mathbb{R}^3\rightarrow\mathbb{R}$, with $$f(x)=\frac{1}{\lvert x\rvert^{2}}$$ which is $$\frac{1}{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}$$ for $x=(x_{1},x_{2},x_{3})$. I now have $$\hat{f}(\xi_{1},\xi_{2},\xi_{3})=\int_{\mathbb{R}^3}e^{-2\pi ix\xi}f(x)dx=\int_{\mathbb{R}^3}e^{-2\pi i(x_{1}\xi_{1}+x_{2}\xi_{2}+x_{3}\xi_{3})}\frac{1}{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}dx_{1}dx_{2}dx_{3}$$
and pretty much that's is from me. Does someone know how to proceed further? Is the fact that $f$ is radial of any help here?
According to Bracewell, the $n\mathrm{-dimensional}$ Fourier Transform, with symmetry in $n$ dimensions, is related to the order $\frac{1}{2}n-1$ Hankel Transform, which yields the symmetrical $n\mathrm{-dimensional}$ Fourier Transform as:
$$F(s) = \dfrac{2\pi}{s^{\frac{1}{2}n-1}}\int_{0}^{\infty}f(r)J_{\frac{1}{2}n -1}(2\pi sr)r^{\frac{1}{2}n}dr$$
which for $n=3$ works out to:
$$F(s) = 4\pi \int_{0}^{\infty}f(r)\mathrm{sinc}(2sr)r^{2}dr$$
where
$$ \mathrm{sinc}(x) \equiv \dfrac{\sin(\pi x)}{\pi x}$$
Applying that transform to your function of interest:
$$\begin{align*}F(s) &= 4\pi \int_{0}^{\infty}\dfrac{1}{r^2}\mathrm{sinc}(2sr)r^{2}dr\\ \\ &= 4\pi \int_{0}^{\infty}\mathrm{sinc}(2sr)dr\\ \\ &= 2\pi \int_{-\infty}^{\infty}\mathrm{sinc}(2sr)dr\\ \\ &=\dfrac{2\pi}{|2s|}\\ \\ &=\dfrac{\pi}{|s|}\\ \end{align*}$$
where I used the result that
$$\int_{-\infty}^{\infty}\mathrm{sinc}(ax)dx = \dfrac{1}{|a|}$$