The correct option is (b).

- Case 1: $x \geq 1$. In this case, $x-1 \geq 0$ and, therefore, $|x-1|=x-1$. The given equation takes the form $(x-1) / x^{2}=6$, or $6 x^{2}-x+1=0$. This equation has no real roots because $\Delta=(-1)^{2}-4 \times 6 \times 1=1-24$ is negative. Therefore, we have no solutions $x$ greater than or equal to 1.
- Case 2: $x<1$. In this case, $x-1<0$ and, therefore, $|x-1|=-(x-1)=1-x$. The given equation takes the form $(1-x) / x^{2}=6$, or $6 x^{2}+x-1=0$. This equation has the roots

$$
x=\frac{-1 \pm \sqrt{1-4 \times 6 \times(-1)}}{2 \times 6}=\frac{-1 \pm \sqrt{25}}{12}=\frac{-1 \pm 5}{12}
$$

that is, $x=-\frac{1}{2}$ and $x=\frac{1}{3}$.

Based on these two cases, we conclude that our equation has only two solutions, $-\frac{1}{2}$ and $\frac{1}{3}$. Therefore, the smallest solution of the equation is $-\frac{1}{2}$.
What is the smallest root of the equation $\frac{|x-1|}{x^{2}}=6$?
(a) $-\frac{1}{3}$
(b) $-\frac{1}{2}$
(c) $\frac{1}{3}$
(d) $\frac{1}{4}$
(e) $\frac{3}{2}$