let $\phi$ be a test function, let $\epsilon > 0$, then :
$$\langle fp(\frac{1}{x^2}),\phi\rangle = \lim_{\epsilon \to 0} [\int_{|x| \geq \epsilon} \frac{\phi(x)}{x^2}dx - 2\frac{\phi(0)}{\epsilon}]$$
show that this defines a distribution
my attempt :
let $K = [-m,m]$ such that $supp\phi \subset K$
let's first rewrite the expression without the limit :
$$\int_{|x| \geq \epsilon} \frac{\phi(x)}{x^2} dx - 2\frac{\phi(0)}{\epsilon} = \int_{|x| \geq \epsilon} \frac{\phi(x) - \phi(0)}{x^2} dx = \int_{m\geq |x| \geq \epsilon} \frac{\phi(x) - \phi(0)}{x^2} dx$$
let's apply a taylor expansion with integral remainder to $\phi$
$$\phi(x) = \phi(0)+x\phi'(0) +x^2\int_0^1\phi''(xt)(1-t)dt = \phi(0)+x\phi'(0) +x^2\psi(x)$$
we also have $|\psi(x)| \leq \sup_{K} |\phi''|$
so putting all these back into the equation we get :
$$\int_{m\geq |x| \geq \epsilon} \frac{\phi(x) - \phi(0)}{x^2} dx = \int_{m\geq |x| \geq \epsilon} \frac{\phi'(0) }{x} dx + \int_{m\geq |x| \geq \epsilon} \psi(x) dx$$
first term is zero because we're integrating an odd function over a symetric interval.
and by the dominated convergence theorem we get :
$$\langle fp(\frac{1}{x^2}),\phi\rangle = \int_{m\geq |x|} \psi(x) dx $$
and
$$|\langle fp(\frac{1}{x^2}),\phi\rangle| \leq 2m \sup_K|\phi''|$$
so it defines a distribution of order $\leq 2$
linearity is obvious so I think I'm done.
is my work all right ?