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

1. Problem Description

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

2. Solution Steps

First, simplify the denominator:

$\frac{4}{y+2} + 1 = \frac{4}{y+2} + \frac{y+2}{y+2} = \frac{4 + y + 2}{y+2} = \frac{y+6}{y+2}$.

So the complex fraction becomes $\frac{\frac{2}{y^2-4}}{\frac{y+6}{y+2}}$.

To divide fractions, we multiply by the reciprocal of the denominator:

$\frac{\frac{2}{y^2-4}}{\frac{y+6}{y+2}} = \frac{2}{y^2-4} \cdot \frac{y+2}{y+6}$.

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

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

So we have $\frac{2}{(y-2)(y+2)} \cdot \frac{y+2}{y+6} = \frac{2(y+2)}{(y-2)(y+2)(y+6)}$.

Now, we can cancel the common factor $(y+2)$ from the numerator and denominator, assuming $y \ne -2$:

$\frac{2(y+2)}{(y-2)(y+2)(y+6)} = \frac{2}{(y-2)(y+6)}$.

Thus, the simplified expression is $\frac{2}{(y-2)(y+6)}$. We can expand the denominator:

$(y-2)(y+6) = y^2 + 6y - 2y - 12 = y^2 + 4y - 12$.

Therefore, the simplified complex fraction is $\frac{2}{y^2 + 4y - 12}$.

3. Final Answer

$\frac{2}{y^2+4y-12}$