Answer
$-\displaystyle \frac{(x-4)^{2}}{4x}$
Work Step by Step
$...=\displaystyle \frac{4-x}{4+x}\div\frac{4x}{x^{2}-16}=\frac{4-x}{4+x}\cdot\frac{x^{2}-16}{4x}$
Factoring,
$ x^{2}-16=(x+4)(x-4)\quad$ (a difference of squares)
$4-x=-(x-4)$
... = $\displaystyle \frac{-(x-4)(x-4)(x+4)}{(x+4)\cdot 4x}$
... cancel the following from both sides of the fraction line:
$(x+4)$
... = $-\displaystyle \frac{(x-4)(x-4)}{4x}$
= $-\displaystyle \frac{(x-4)^{2}}{4x}$
