Answer
$\left( -\dfrac{22}{3},\dfrac{14}{3} \right)$
Work Step by Step
Since for any $a\gt0$, $|x|\lt a$ implies $-a\lt x\lt a$, then the solution to the given inequality, $
\left| \dfrac{1}{2}x+\dfrac{2}{3} \right|\lt3
,$ is
\begin{array}{l}\require{cancel}
-3\lt \dfrac{1}{2}x+\dfrac{2}{3}\lt3
\\\\
6\cdot(-3)\lt 6\cdot\left( \dfrac{1}{2}x+\dfrac{2}{3} \right)\lt6\cdot(3)
\\\\
-18\lt 3x+4\lt18
\\\\
-18-4\lt 3x+4-4\lt18-4
\\\\
-22\lt 3x\lt14
\\\\
-\dfrac{22}{3}\lt \dfrac{3x}{3}\lt\dfrac{14}{3}
\\\\
-\dfrac{22}{3}\lt x\lt\dfrac{14}{3}
.\end{array}
Hence, the solution set is the interval $
\left( -\dfrac{22}{3},\dfrac{14}{3} \right)
.$