#### Answer

$\dfrac{1}{\sqrt{15}+3}$

#### Work Step by Step

Multiplying by the conjugate of the numerator, the rationalized-numerator form of the given expression, $
\dfrac{\sqrt{15}-3}{6}
,$ is
\begin{array}{l}\require{cancel}
\dfrac{\sqrt{15}-3}{6}\cdot\dfrac{\sqrt{15}+3}{\sqrt{15}+3}
\\\\=
\dfrac{(\sqrt{15})^2-(3)^2}{6(\sqrt{15})+6(3)}
\\\\=
\dfrac{15-9}{6\sqrt{15}+18}
\\\\=
\dfrac{6}{6(\sqrt{15}+3)}
\\\\=
\dfrac{\cancel{6}}{\cancel{6}(\sqrt{15}+3)}
\\\\=
\dfrac{1}{\sqrt{15}+3}
.\end{array}