We are asked to simplify the complex fraction $\frac{\frac{7}{y^2 - 4}}{\frac{7}{y+2} + 1}$.

1. Problem Description

We are asked to simplify the complex fraction $\frac{\frac{7}{y^2 - 4}}{\frac{7}{y+2} + 1}$.

2. Solution Steps

First, we simplify the denominator of the complex fraction:

$\frac{7}{y+2} + 1 = \frac{7}{y+2} + \frac{y+2}{y+2} = \frac{7 + y + 2}{y+2} = \frac{y+9}{y+2}$.

Then, we can rewrite the complex fraction as:

$\frac{\frac{7}{y^2 - 4}}{\frac{y+9}{y+2}} = \frac{7}{y^2 - 4} \div \frac{y+9}{y+2} = \frac{7}{y^2 - 4} \cdot \frac{y+2}{y+9}$.

Next, we can factor the denominator $y^2 - 4$ as a difference of squares:

$y^2 - 4 = (y-2)(y+2)$.

So we have:

$\frac{7}{(y-2)(y+2)} \cdot \frac{y+2}{y+9} = \frac{7(y+2)}{(y-2)(y+2)(y+9)}$.

We can cancel the $(y+2)$ term from the numerator and denominator, assuming $y \neq -2$:

$\frac{7}{(y-2)(y+9)}$.

Finally, we can expand the denominator:

$(y-2)(y+9) = y^2 + 9y - 2y - 18 = y^2 + 7y - 18$.

Thus, the simplified fraction is:

$\frac{7}{y^2 + 7y - 18}$.

3. Final Answer

$\frac{7}{y^2+7y-18}$