= Solution
Choose $R>0$ with $\operatorname{supp}\phi\subset[-R,R]$. Evenness of $f$ changes the symmetrically truncated integral into
$$
\int_{|x|>\varepsilon}\frac{f(x)}x\phi(x)\,dx
=\int_\varepsilon^R f(x)\frac{\phi(x)-\phi(-x)}x\,dx.
$$
The <mean value theorem> gives $|\phi(x)-\phi(-x)|\leq2x\|\phi'\|_\infty$. Since $f$ is continuous, it is bounded on $[0,R]$. Thus the <Cauchy principal value> exists and
$$
\boxed{\left\langle\operatorname{pv}\frac{f(x)}x,\phi\right\rangle
=\int_0^R f(x)\frac{\phi(x)-\phi(-x)}x\,dx.}
$$
This expression is linear in the <test function> and satisfies
$$
\left|\left\langle\operatorname{pv}\frac{f(x)}x,\phi\right\rangle\right|
\leq2R\max_{[0,R]}|f|\,\|\phi'\|_\infty.
$$
For each fixed <compact support>, choose one $R$ once and for all. The estimate proves that the <even-weight principal-value distribution> has <order of a distribution> at most one.
If $\phi(0)=0$, the <mean value theorem> also gives $|\phi(x)|\leq|x|\|\phi'\|_\infty$. Hence $f(x)\phi(x)/x$ is absolutely integrable near zero and has <compact support>. The truncated integrals tend to its ordinary <Lebesgue integral>, so
$$
\boxed{\left\langle\operatorname{pv}\frac{f(x)}x,\phi\right\rangle
=\int_{\mathbb R}\frac{f(x)}x\phi(x)\,dx\quad\text{if }\phi(0)=0.}
$$
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