Answer
$\dfrac{y+3}{y+4}$
Work Step by Step
The given expression, $
\dfrac{y^2+y-2}{y^2+3y-4}\div\dfrac{y^2+3y+2}{y^2+4y+3}
,$ simplifies to
\begin{array}{l}\require{cancel}
\dfrac{y^2+y-2}{y^2+3y-4}\cdot\dfrac{y^2+4y+3}{y^2+3y+2}
\\\\=
\dfrac{(y+2)(y-1)}{(y+4)(y-1)}\cdot\dfrac{(y+3)(y+1)}{(y+2)(y+1)}
\\\\=
\dfrac{(\cancel{y+2})(\cancel{y-1})}{(y+4)(\cancel{y-1})}\cdot\dfrac{(y+3)(\cancel{y+1})}{(\cancel{y+2})(\cancel{y+1})}
\\\\=
\dfrac{y+3}{y+4}
.\end{array}
