Answer
$\left( -\infty,-3 \right)\cup(0,\infty)$
Work Step by Step
Since for any $a\gt0$, $|x|\gt a$ implies $x\gt a$ OR $x\lt -a$, then the given inequality, $
\left| \dfrac{2}{3}x+1 \right|\gt1
,$ is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{2}{3}x+1\gt1
\\\\
\dfrac{2}{3}x\gt1-1
\\\\
\dfrac{2}{3}x\gt0
\\\\
\dfrac{3}{2}\cdot\dfrac{2}{3}x\gt0\cdot\dfrac{3}{2}
\\\\
x\gt0
,\\\\\text{OR}\\\\
\dfrac{2}{3}x+1\lt-1
\\\\
\dfrac{2}{3}x\lt-1-1
\\\\
\dfrac{2}{3}x\lt-2
\\\\
\dfrac{3}{2}\cdot\dfrac{2}{3}x\lt-2\cdot\dfrac{3}{2}
\\\\
x\lt-3
.\end{array}
Hence, the solution set is $
\left( -\infty,-3 \right)\cup(0,\infty)
.$
See graph below.