I computed first derivation and I get that $$\langle(\mathcal{P}\frac{1}{x})', \varphi\rangle = v.p. \int_{ \mathbb{R}} \frac{\varphi(0) - \varphi(x)}{x^2} dx$$
In order to get second derivative we use $\left<g'', \varphi \right> = \left<g, \varphi''\right> $ but I stuck there since I have just a lot of equations that gives us nothing.
Any help would be great!
Let $$ \operatorname{pv}\int f(x) \ dx = \lim_{\epsilon\to 0} \int_{|x|>\epsilon} f(x) \ dx. $$
We define the distribution $\operatorname{pv}\frac{1}{x}$ by $$ \langle \operatorname{pv}\frac{1}{x}, \varphi \rangle = \operatorname{pv} \int \frac{\varphi(x)}{x} \ dx . $$
For its derivative we get $$ \langle \left(\operatorname{pv}\frac{1}{x}\right)', \varphi \rangle = -\langle \operatorname{pv}\frac{1}{x}, \varphi' \rangle = -\operatorname{pv}\int \frac{\varphi'(x)}{x} \ dx = -\lim_{\epsilon\to 0} \int_{|x|>\epsilon} \frac{\varphi'(x)}{x} \ dx \\ = -\lim_{\epsilon\to 0} \left( \left[ \frac{\varphi(x)-\varphi(0)}{x} \right]_{-\infty}^{-\epsilon} + \left[ \frac{\varphi(x)-\varphi(0)}{x} \right]_{\epsilon}^{\infty} + \int_{|x|>\epsilon} \frac{\varphi(x)-\varphi(0)}{x^2} \ dx \right)\\ = -\lim_{\epsilon\to 0} \int_{|x|>\epsilon} \frac{\varphi(x)-\varphi(0)}{x^2} \ dx = -\operatorname{pv}\int \frac{\varphi(x)-\varphi(0)}{x^2} \ dx . $$
For the second derivative we then get $$ \langle \left(\operatorname{pv}\frac{1}{x}\right)'', \varphi \rangle = -\langle \left(\operatorname{pv}\frac{1}{x}\right)', \varphi' \rangle = \operatorname{pv}\int \frac{\varphi'(x)-\varphi'(0)}{x^2} \ dx = \lim_{\epsilon\to 0} \int_{|x|>\epsilon} \frac{\varphi'(x)-\varphi'(0)}{x^2} \ dx \\ = \lim_{\epsilon\to 0} \left( \left[ \frac{\varphi(x)-\varphi'(0)x-\varphi(0)}{x^2} \right]_{-\infty}^{-\epsilon} + \left[ \frac{\varphi(x)-\varphi'(0)x-\varphi(0)}{x^2} \right]_{\epsilon}^{\infty} \\ + 2 \int_{|x|>\epsilon} \frac{\varphi(x)-\varphi'(0)x-\varphi(0)}{x^3} \ dx \right) \\ = 2\operatorname{pv}\int \frac{\varphi(x)-\varphi'(0)x-\varphi(0)}{x^3} \ dx. $$