Решение:

$$\frac{2}{x-4} - \frac{x+8}{x^2 - 16} - \frac{1}{x} = \frac{2}{x-4} - \frac{x+8}{(x-4)(x+4)} - \frac{1}{x} = \frac{2x(x+4) - x(x+8) - (x-4)(x+4)}{x(x-4)(x+4)} = $$ $$\frac{2x^2+8x - x^2 -8x - (x^2 - 16)}{x(x-4)(x+4)} = \frac{2x^2 + 8x - x^2 - 8x - x^2 + 16}{x(x-4)(x+4)} = \frac{16}{x(x-4)(x+4)} = \frac{16}{x(x^2 - 16)}$$

Ответ: $$\frac{16}{x(x^2 - 16)}$$