Answer
$$y'=\frac{16\sin2x(1-\cos2x)^3}{(1+\cos2x)^5}$$
Work Step by Step
$$y'=\left(\left(\frac{1-\cos2x}{1+\cos2x}\right)^4\right)'=
4\left(\frac{1-\cos2x}{1+\cos2x}\right)^3\left(\frac{1-\cos2x}{1+\cos2x}\right)'=
4\left(\frac{1-\cos2x}{1+\cos2x}\right)^3\frac{(1-\cos2x)'(1+\cos2x)-(1-\cos2x)(1+\cos2x)'}{(1+\cos2x)^2}=
4\left(\frac{1-\cos2x}{1+\cos2x}\right)^3\frac{-(-\sin2x)(2x)'(1+\cos2x)-(1-\cos2x)(-\sin2x)(2x)'}{(1+\cos2x)^2}=4\left(\frac{1-\cos2x}{1+\cos2x}\right)^3\frac{\sin2x\cdot2(1+\cos2x)+\sin2x(1-\cos2x)\cdot2}{(1+\cos2x)^2}=
4\left(\frac{1-\cos2x}{1+\cos2x}\right)^3\frac{2\sin2x+2\sin2x\cos2x+2\sin2x-2\sin2x\cos2x}{(1+\cos2x)^2}=
4\left(\frac{1-\cos2x}{1+\cos2x}\right)^3\frac{4\sin2x}{(1+\cos2x)^2}=
\frac{16\sin2x(1-\cos2x)^3}{(1+\cos2x)^5}$$
