If $x=\sin \theta, y=\sin ^3 \theta$, then $\frac{d^2 y}{d x^2}$ at $\theta=\frac{\pi}{6}$ is

If $y=\tan ^{-1}\left[\frac{4 \sin 2 x}{\cos 2 x-6 \sin ^2 x}\right]$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ at $x=0$ is

If $x=a \sin 2 t(1+\cos 2 t), y=b \cos 2 t(1-\cos 2 t)$ then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ is equal to