$f(x)={{\left( \frac{\sin 2x}{1+\cos 2x} \right)}^{\frac{1}{3}}}={{(\frac{2\sin x\cos x}{2{{\cos }^{2}}x})}^{\frac{1}{3}}}={{(\tan x)}^{\frac{1}{3}}}$ $\Rightarrow {f}'(x)=\frac{1}{3}{{(\tan x)}^{-\frac{2}{3}}}\times (1+{{\tan }^{2}}x)\Rightarrow {f}'(\frac{\pi }{4})=\frac{1}{3}\times {{(1)}^{-\frac{2}{3}}}\times (1+1)=\frac{2}{3}$