Rationalize the denominator:

$\dfrac{\sqrt{2}}{2\sqrt{3}}$

Rationalizing the denominator, $\Rightarrow \dfrac{\sqrt{2}}{2\sqrt{3}} \times \dfrac{2\sqrt{3}}{2\sqrt{3}} \\[1em] \Rightarrow \dfrac{2(\sqrt{6})}{(2\sqrt{3})^2} \\[1em] \Rightarrow \dfrac{2\sqrt{6}}{(4 \times 3)} \\[1em] \Rightarrow \dfrac{2\sqrt{6}}{12} \\[1em] \Rightarrow \dfrac{\sqrt{6}}{6}.$

Hence, on rationalizing = $\dfrac{\sqrt{2}}{2\sqrt{3}} = \dfrac{\sqrt{6}}{6}$.