If $x = 3\sin t - \sin 3t,y = 3\cos t - \cos 3t,$ then find $\frac{{dy}}{{dx}}$ at $t = \frac{\pi }{3}$

and $y = 3\cos t - \cos 3t$
therefore,$\frac{{dx}}{{dt}} = 3 \cdot \frac{d}{{dt}}\sin t - \frac{d}{{dt}}\sin 3t$

$= 3\cos t - \cos 3t \cdot \frac{d}{{dt}}3t = 3\cos t - 3\cos 3t$ ……..(i)

and $\frac{{dy}}{{dt}} = 3 \cdot \frac{d}{{dt}}\cos t - \frac{d}{{dt}}\cos 3t$
$= - 3\sin t + \sin 3t \cdot \frac{d}{{dt}}3t$
$\frac{{dy}}{{dt}} = 3\sin 3t - 3t\sin t$ ……(ii)
therefore,$\frac{{dy}}{{dx}} = \frac{{dy/dt}}{{dx/dt}} = \frac{{3(\sin 3t - \sin t)}}{{3(\cos t - \cos 3t)}}$

Now, ${\left( {\frac{{dy}}{{dx}}} \right)_{t = \pi /3}} = \frac{{\sin \frac{{3\pi }}{3} - \sin \frac{\pi }{3}}}{{\left( {\cos \frac{\pi }{3} - \cos 3\frac{\pi }{3}} \right)}} = \frac{{0 - \sqrt 3 /2}}{{\frac{1}{2} - ( - 1)}}$

$= \frac{{ - \sqrt 3 /2}}{{3/2}} = \frac{{ - \sqrt 3 }}{3} = \frac{{ - 1}}{{\sqrt 3 }}$

$\Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)_{t = \pi /3}} = \frac{{ - 1}}{{\sqrt 3 }}$