Answer
$ \left( -\infty, -12 \right)\cup\left( 0,\infty \right) $
Work Step by Step
Since for any $a\gt0$, $|x|\gt a$ implies $x\gt a\text{ or } x\lt-a$, then the expression, $ \left| \dfrac{x+6}{3} \right|\gt2 ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{x+6}{3}\gt2 \\\\ x+6\gt2(3) \\\\ x+6\gt6 \\\\ x\gt6-6 \\\\ x\gt0 ,\\\\\text{ OR }\\\\ \dfrac{x+6}{3}\lt-2 \\\\ x+6\lt-2(3) \\\\ x+6\lt-6 \\\\ x\lt-6-6 \\\\ x\lt-12 .\end{array} Hence, the solution set is $ \left( -\infty, -12 \right)\cup\left( 0,\infty \right) .$