How understand this Fourier transform: $\widehat{\frac{1}{x^3}}=c \xi^{2} sgn(\xi)$ ?
The function $x\mapsto\frac{1}{x}$ is not locally integrable, so it does not define a distribution through simple integration against test functions. A typical way of defining a distribution related to $\frac{1}{x}$ is $p.v.\frac{1}{x}$ which is defined through the pairing $$\langle p.v.\frac{1}{x}, \varphi\rangle=p.v.\int \frac{\varphi}{x}:=\lim_{\epsilon\rightarrow 0^{+}}\int_{|x|\geq \epsilon}\frac{\varphi(x)}{x}dx.$$ The same remark applies of course to $\frac{1}{x^3}$.
So, the first question is how justify the Fourier transform of $\frac{1}{x}$ if it is not even a distribution ?
Second, pretending that there is not a too bad singularity at the origin (which is my first question), and integrating by parts twice, we have $$\int \frac{e^{\dot{\imath}x\xi}}{x^3}dx = \frac{\dot{\imath}\xi}{2}\int \frac{e^{\dot{\imath}x\xi}}{x^2}dx=-\frac{\xi^{2}}{2}\int \frac{e^{\dot{\imath}x\xi}}{x}dx$$ Now, the integral $\int\frac{\sin{\xi x}}{x}$ does converge to $\pi sgn{(\xi)}$, while the integral $\int\frac{\cos{x}}{x}$ diverges, but converges in principal value to 0.
The Fourier transform of $\operatorname{pv} \frac{1}{x}$ is $-i\pi \, \operatorname{sign}(\xi).$ (See Fourier transform of the distribution PV $\left( \frac{1}{x} \right)$)
Now, $\operatorname{pv} \frac{1}{x^3} = \frac{1}{2} (\operatorname{pv}\frac{1}{x})''$ so applying the derivative rule $\mathcal{F}\{f'\} = i\xi \mathcal{F}\{f\}$ gives $$\mathcal{F}\{\operatorname{pv} \frac{1}{x^3}\} = (i\xi)^2 \mathcal{F}\{\operatorname{pv} \frac{1}{x}\} = \frac{i\pi}{2}\xi^2 \operatorname{sign}(\xi).$$