Answer
$$\frac{dy}{dx}=\frac{\cos x-3\sin x\tan(3x+1)}{\sec(3x+1)}$$
Work Step by Step
$$\frac{dy}{dx}=\frac{d}{dx}\Big(\frac{\sin x}{\sec(3x+1)}\Big)=
\frac{(\sin x)'\sec(3x+1)-\sin x(\sec(3x+1))'}{\sec^2(3x+1)}=
\frac{\cos x\sec(3x+1)-\sin x\sec(3x+1)\tan(3x+1)(3x+1)'}{\sec^2(3x+1)}=
\frac{\cos x-\sin x\tan(3x+1)\cdot3}{\sec(3x+1)}=
\frac{\cos x-3\sin x\tan(3x+1)}{\sec(3x+1)}$$
