Answer
$\frac{y^3-9y+1}{y-3}$
Work Step by Step
Step 1. Perform operations to the numerator:
$y+\frac{1}{y^2-9}=\frac{y^3-9y+1}{y^2-9}=\frac{y^3-9y+1}{(y+3)(y-3)}$
Step 2. Use the above result in the original expression, we have:
$\frac{y+\frac{1}{y^2-9}}{\frac{1}{y+3}}=\frac{\frac{y^3-9y+1}{(y+3)(y-3)}}{\frac{1}{y+3}}=\frac{y^3-9y+1}{(y+3)(y-3)}\cdot \frac{y+3}{1}=\frac{y^3-9y+1}{y-3}$
