Answer
$\dfrac{3\sqrt[]{y}+6}{y-4}$
Work Step by Step
Multiplying by the conjugate of the denominator, the rationalized-denominator form of the given expression, $
\dfrac{3}{\sqrt[]{y}-2}
,$ is
\begin{array}{l}\require{cancel}
\dfrac{3}{\sqrt[]{y}-2}\cdot\dfrac{\sqrt[]{y}+2}{\sqrt[]{y}+2}
\\\\=
\dfrac{3\sqrt[]{y}+6}{(\sqrt[]{y})^2-(2)^2}
\\\\=
\dfrac{3\sqrt[]{y}+6}{y-4}
.\end{array}
Note that all variables are assumed to have positive real numbers.