Answer
$x\displaystyle \in\left\{\frac{1}{3},\frac{1}{2},1\right\}$
Work Step by Step
See The Rational Zero Theorem:
If $\frac{p}{q}$ is a zero of the polynomial $f(x) $with integer coefficients, then $p$ is a factor of the constant term, $a_{0}$, and $q$ is a factor of the leading coefficient, $a_{n}$.
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$6x^{3}-11x^{2}+6x-1=0$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1 $
$q:\qquad \pm 1, \pm 2, \pm 3, \pm 6$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm\frac{1}{2},\pm\frac{1}{3}, \pm\frac{1}{6}$
b. Try for $x=1:$
$\begin{array}{lllll}
\underline{1}| & 6 & -11 & 6 & -1\\
& & 6 & -5 & 1\\
& -- & -- & -- & --\\
& 6 & -5 & 1 & |\underline{0}
\end{array}$
$1$ is a zero,
$(x-1)(6x^{2} -5x+1)$
c. Factorize the trinomial factor $(6x^{2} -5x+1)$
$6x^{2} -5x+1=6x^{2} -3x-2x+1 \quad$...factor in pairs ...
$=3x(2x-1)-1(2x-1)=(2x-1)(3x-1)$
$(x-1)(2x-1)(3x-1)=0$
$x\displaystyle \in\left\{\frac{1}{3},\frac{1}{2},1\right\}$
