Answer
$\left[ -1,-\dfrac{1}{2} \right]$
Work Step by Step
Since for any $a\gt0$, $|x|\le a$ implies $-a\le x \le a$ then the solution to the given inequality, $
\left| \dfrac{2}{3}x+\dfrac{1}{2} \right|\le \dfrac{1}{6}
,$ is
\begin{array}{l}\require{cancel}
-\dfrac{1}{6}\le \dfrac{2}{3}x+\dfrac{1}{2} \le \dfrac{1}{6}
\\\\
6\left( -\dfrac{1}{6} \right)\le 6\left( \dfrac{2}{3}x+\dfrac{1}{2} \right) \le 6\left( \dfrac{1}{6} \right)
\\\\
-1\le 4x+3 \le 1
\\\\
-1-3\le 4x+3-3 \le 1-3
\\\\
-4\le 4x \le -2
\\\\
-\dfrac{4}{4}\le \dfrac{4x}{4} \le \dfrac{-2}{4}
\\\\
-1\le x \le -\dfrac{1}{2}
.\end{array}
Hence, the solution set is the interval $
\left[ -1,-\dfrac{1}{2} \right]
.$