Answer
$(-4, -2) U (2, 4)$
Work Step by Step
$\frac{x^2 - 16}{x^4 - 16} < 0$
$\frac{(x-4)(x+4)}{(x^2 - 4) (x^2 + 4)} < 0$
$\frac{(x-4)(x+4)}{(x-2)(x+2)(x^2+4)} < 0$
Find the zeros of the expressions in the numerator AND the denominator
$x = -2, 2, 4, -4$
Test numbers in between those zero values to determine if the function is negative or positive
$(-∞, -4)$ $\frac{(-)(-))}{(-)(-)(+)} = (+)$
$(-4, -2])$ $\frac{(-)(+)}{(-)(-)(+)} = (-)$
$(-2, 2)$ $\frac{(-)(+)}{(-)(+)(+)} = (+)$
$(2, 4)$ $\frac{(-)(+)}{(+)(+)(+)} = (-)$
$(4, ∞)$ $\frac{(+)(+)}{(+)(+)(+))} = (+)$
Thus the solution is $(-4, -2) U (2, 4)$