Answer
$$y'=\frac{2 \sec 2 \theta(\tan 2 \theta-1)}{(1+\tan 2 \theta)^{2}}$$
Work Step by Step
Given
$$y=\frac{\sec2\theta }{1+\tan 2\theta }$$
Then
\begin{align*}
y^{\prime} &=\frac{(1+\tan 2 \theta)(\sec 2 \theta \tan 2 \theta \cdot 2)-(\sec 2 \theta)\left(\sec ^{2} 2 \theta \cdot 2\right)}{(1+\tan 2 \theta)^{2}}\\
&=\frac{2 \sec 2 \theta\left[(1+\tan 2 \theta) \tan 2 \theta-\sec ^{2} 2 \theta\right]}{(1+\tan 2 \theta)^{2}} \\ &=\frac{2 \sec 2 \theta\left(\tan 2 \theta+\tan ^{2} 2 \theta-\sec ^{2} 2 \theta\right)}{(1+\tan 2 \theta)^{2}} \ \text{Using}\quad\left[1+\tan ^{2} x=\sec ^{2} x\right]\\
&=\frac{2 \sec 2 \theta(\tan 2 \theta-1)}{(1+\tan 2 \theta)^{2}}
\end{align*}