Answer
$\dfrac{1}{y^2-3y+9}$
Work Step by Step
The given expression, $
\dfrac{\dfrac{1}{y}+\dfrac{3}{y^2}}{y+\dfrac{27}{y^2}}
,$ simplifies to
\begin{array}{l}\require{cancel}
\dfrac{\dfrac{y(1)+1(3)}{y^2}}{\dfrac{y^2(y)+1(27)}{y^2}}
\\\\=
\dfrac{\dfrac{y+3}{y^2}}{\dfrac{y^3+27}{y^2}}
\\\\=
\dfrac{y+3}{y^2}\div{\dfrac{y^3+27}{y^2}}
\\\\=
\dfrac{y+3}{y^2}\cdot{\dfrac{y^2}{y^3+27}}
\\\\=
\dfrac{y+3}{y^2}\cdot{\dfrac{y^2}{(y+3)(y^2-3y+9)}}
\\\\=
\dfrac{\cancel{y+3}}{\cancel{y^2}}\cdot{\dfrac{\cancel{y^2}}{(\cancel{y+3})(y^2-3y+9)}}
\\\\=
\dfrac{1}{y^2-3y+9}
.\end{array}
