The problem requires us to simplify the expression $\frac{8}{4y-y^3} + \frac{1}{y+2} - \frac{1}{y}$.

1. Problem Description

The problem requires us to simplify the expression $\frac{8}{4y-y^3} + \frac{1}{y+2} - \frac{1}{y}$.

2. Solution Steps

First, factor the denominator of the first term:

$4y - y^3 = y(4-y^2) = y(2-y)(2+y) = -y(y-2)(y+2)$

The expression becomes:

$\frac{8}{-y(y-2)(y+2)} + \frac{1}{y+2} - \frac{1}{y}$

Now, find a common denominator, which is $-y(y-2)(y+2)$. Rewrite each term with the common denominator:

$\frac{8}{-y(y-2)(y+2)} + \frac{-y(y-2)}{(y+2)(-y(y-2))} - \frac{-(y-2)(y+2)}{y(-1)(y-2)(y+2)}$

$\frac{8}{-y(y-2)(y+2)} + \frac{-y(y-2)}{-y(y-2)(y+2)} - \frac{-(y^2-4)}{-y(y-2)(y+2)}$

Combine the numerators:

$\frac{8 -y(y-2) + (y^2-4)}{-y(y-2)(y+2)} = \frac{8 - y^2 + 2y + y^2 - 4}{-y(y-2)(y+2)}$

Simplify the numerator:

$\frac{4 + 2y}{-y(y-2)(y+2)} = \frac{2(2+y)}{-y(y-2)(y+2)}$

Cancel the common factor $(y+2)$:

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

3. Final Answer

$\frac{2}{2y-y^2}$