Answer
$\dfrac{3\sqrt{2}+\sqrt{6}}{6}$
Work Step by Step
Multiplying by the conjugate of the denominator and using $(a+b)(a-b)=a^2-b^2$, the simplifed form of given expression, $
\dfrac{2}{3\sqrt{2}-\sqrt{6}}
,$ is
\begin{array}{l}\require{cancel}
\dfrac{2}{3\sqrt{2}-\sqrt{6}}\cdot\dfrac{3\sqrt{2}+\sqrt{6}}{3\sqrt{2}+\sqrt{6}}
\\\\=
\dfrac{2(3\sqrt{2}+\sqrt{6})}{(3\sqrt{2})^2-(\sqrt{6})^2}
\\\\=
\dfrac{2(3\sqrt{2}+\sqrt{6})}{9(2)-6}
\\\\=
\dfrac{2(3\sqrt{2}+\sqrt{6})}{12}
\\\\=
\dfrac{\cancel{2}(3\sqrt{2}+\sqrt{6})}{\cancel{2}(6)}
\\\\=
\dfrac{3\sqrt{2}+\sqrt{6}}{6}
.\end{array}
