Answer
$$\frac{dy}{dx}=0$$
Work Step by Step
$$y=(\sec x+\tan x)(\sec x-\tan x)$$
$$\frac{dy}{dx}=\frac{d}{dx}(\sec x+\tan x)(\sec x-\tan x)$$ $$=(\sec x+\tan x)'(\sec x-\tan x)+(\sec x+\tan x)(\sec x-\tan x)'$$ $$=\Big((\sec x)'+(\tan x)'\Big)(\sec x-\tan x)+(\sec x+\tan x)\Big((\sec x)'-(\tan x)'\Big)$$ $$=(\sec x\tan x+\sec^2x)(\sec x-\tan x)+(\sec x+\tan x)(\sec x\tan x-\sec^2x)$$ $$=\sec x(\sec x+\tan x)(\sec x-\tan x)+(\sec x+\tan x)\sec x(\tan x-\sec x)$$ $$=\sec x(\sec x+\tan x)(\sec x-\tan x)-\sec x(\sec x+\tan x)(\sec x-\tan x)$$ $$=0$$
