Using
$$F[\theta] = \pi \delta(\xi) + i p.v. \frac{1}{\xi}$$
and
$$F[\theta(-x)] = \pi \delta(\xi) - i p.v. \frac{1}{\xi}$$
and
$$p.v. \frac{1}{\xi^{2}} = - (p.v. \frac{1}{\xi})'$$
show the Fourier transforms
$$F[p.v. \frac{1}{x^{2}}] = -\pi |\xi|, \quad F[|x|] = -2 p.v. \frac{1}{\xi^{2}}.$$
Where $\theta$ is the Heaviside function and $F[]$ is the Fourier transform.
Edit: Not sure why this was downvoted for showing clarity but I copied the problem statement from the textbook.
The first: $$ \mathcal{F}\{ \frac{1}{x^2} \} = \mathcal{F}\{ -(\frac{1}{x})' \} = - \mathcal{F}\{ (\frac{1}{x})' \} = -i\xi \, \mathcal{F}\{ \frac{1}{x} \} = -i\xi \, \mathcal{F}\{ \frac{1}{2i}(\mathcal{F}\{\theta(x)\}-\mathcal{F}\{\theta(-x)\}) \} \\ = -\frac{1}{2}\xi \, \mathcal{F}\{ (\mathcal{F}\{\theta(x)\}-\mathcal{F}\{\theta(-x)\}) \} = -\frac{1}{2}\xi \, (\mathcal{F}\{\mathcal{F}\{\theta(x)\}\} - \mathcal{F}\{\mathcal{F}\{\theta(-x)\}\} ) \\ = -\frac{1}{2}\xi \, (2\pi\,\theta(-\xi) - 2\pi\,\theta(\xi) ) = \pi\xi \, (\theta(\xi) - \theta(-\xi) ) = \pi\xi \operatorname{sign}(\xi) = \pi |\xi|. $$
The second: $$ \mathcal{F}\{|x|\} = \frac{1}{\pi} \mathcal{F}\{\pi|x|\} = \frac{1}{\pi} \mathcal{F}\{\mathcal{F}\{\frac{1}{\xi^2}\}\} = \frac{1}{\pi} \, 2\pi\frac{1}{(-\xi)^2} = \frac{2}{\xi^2}. $$ I don't get the minus sign that you give, however. Have I made an error, or have you made an error, or has the teacher made an error?