Find the equation

finding slope of the tangent by differentiating $x$ and $y$ with respect to theta

$\frac{\mathrm{dx}}{\mathrm{d} \theta}=-3 \sin \theta+3 \cos ^{2} \theta \sin \theta$

$\frac{\mathrm{dy}}{\mathrm{d} \theta}=3 \cos \theta-3 \sin ^{2} \theta \cos \theta$

Now dividing $\frac{d y}{d \theta}$ and $\frac{d x}{d \theta}$ to obtain the slope of tangent

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{3 \cos \theta-3 \sin ^{2} \theta \cos \theta}{-3 \sin \theta+3 \cos ^{2} \theta \sin \theta}=-\tan ^{3} \theta$

$\frac{d y}{d x}=\frac{3 \cos \theta-3 \sin ^{2} \theta \cos \theta}{-3 \sin \theta+3 \cos ^{2} \theta \sin \theta}=-\tan ^{3} \theta$

$\mathrm{m}$ (tangent) at theta is $-\tan ^{3} \theta$

normal is perpendicular to tangent so, $m_{1} m_{2}=-1$

$\mathrm{m}$ (normal) at theta is $\cot ^{3} \theta$

equation of tangent is given by $y-y_{1}=m($ tangent $)\left(x-x_{1}\right)$

$y-3 \sin \theta+\sin ^{3} \theta=-\tan ^{3} \theta\left(x-3 \cos \theta+3 \cos ^{3} \theta\right)$

equation of normal is given by $y-y_{1}=m($ normal $)\left(x-x_{1}\right)$

$y-3 \sin \theta+\sin ^{3} \theta=\cot ^{3} \theta\left(x-3 \cos \theta+3 \cos ^{3} \theta\right)$