Answer
$$y'=\frac{-2\sec^2x}{(1+\tan x)^3}$$
$$y''=\frac{2(2+\cos2x-\sin2x)}{(\cos x+\sin x)^4}$$
Work Step by Step
$$y'=\left(\frac{1}{(1+\tan x)^2}\right)'=((1+\tan x)^{-2})'=-2(1+\tan x)^{-3}(1+\tan x)'=\frac{-2}{(1+\tan x)^3}(0+\frac{1}{\cos^2x})=\frac{-2\sec^2x}{(1+\tan x)^3}$$
$$y''=(y')'=\left(\frac{-2\sec^2x}{(1+\tan x)^3}\right)'=
\frac{(-2\sec^2x)'(1+\tan x)^3+2\sec^2x((1+\tan x)^3)'}{(1+\tan x)^6}=
\frac{-2\cdot2\sec x(\sec x)'(1+\tan x)^3+2\sec^2x\cdot3(1+\tan x)^2(1+\tan x)'}{(1+\tan x)^6}=
\frac{-4\sec x\sec x\tan x(1+\tan x)^3+6\sec^2x(1+\tan x)^2\sec^2x}{(1+\tan x)^6}=
\frac{2\sec^2x(1+\tan x)^2(-2\tan x(1+\tan x)+3\sec^2x)}{(1+\tan x)^6}=
\frac{2\sec^2x(3\sec^2x-2\tan x-2\tan^2x)}{(1+\tan x)^4}=
\frac{2\sec^2x(3\sec^2x-2\tan x-2\tan^2x)}{(1+\tan x)^4}\cdot\frac{\cos^4x}{\cos^4x}=
\frac{2(2+\cos2x-\sin2x)}{(\cos x+\sin x)^4}$$
