$y={\cos }^{-1}\left(\tan hx\right)+\sin h\left(\sin 6x\right),$

$\Rightarrow \frac{dy}{dx}=\frac{-1}{\sqrt{1-{\tan }^{2}hx}}\frac{d}{dx}\left(\tan hx\right)+\cos h\left(\sin 6x\right).\frac{d}{dx}\left(\sin 6x\right)$

$\Rightarrow \frac{dy}{dx}=\frac{-1}{\sqrt{1-{\tan }^{2}hx}}{\sec }^{2}hx+\cos h\left(\sin 6x\right).\cos 6x.6$

$\Rightarrow \frac{dy}{dx}=\frac{-1}{\sec hx}{\sec }^{2}hx+\cos h\left(\sin 6x\right).\cos 6x.6$

$\therefore \frac{dy}{dx}=\frac{-1}{\cos hx}+\cos h\left(\sin 6x\right).\cos 6x.6$

$=\frac{-1}{\cos hx}+6\cos 6x\cos h\left(\sin 6x\right)$

Hence, option $1$ is correct.