The expression $y^{2}-9$ can be factorized as $(y-3)(y+3)$, $y^{2}-3y$ can be factorized as $y(y-3)$, and $y^{2}-y-6$ can be factorized as $(y-3)(y+2)$.
So, the given expression becomes:
$$
\frac{(y-3)(y+3)}{y^{2}} \cdot \frac{y(y-3)}{(y-3)(y+2)}
$$
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