#### Answer

The Derivative is:
$y'=0$

#### Work Step by Step

$y=(\sec x+\tan x)(\sec x-\tan x)$
Applying the notable product $(a+b)(a-b)=a^2-b^2$
$y=\sec^2 x-\tan^2 x$
Applying Derivative rules:
$y'=\frac{d}{dx}(\sec x\sec x)-\frac{d}{dx}(\tan x\tan x)$
Apply the product rule:
$y'=((\sec x\tan x)(\sec x)+(\sec x)(\sec x\tan x))-((\sec^2 x)(\tan x)+(\tan x)(\sec^2 x))$
$y'=2\sec^2 x\tan x - 2\sec^2 x\tan x$
$y'=0$