Answer
The equation of the tangent line at $\theta =\frac{\pi }{3}$ is
$y=-\sqrt{3} x+\frac{\sqrt{3}}{2}$.
Work Step by Step
Using Eq. (8) the slope of the tangent line is
$\frac{dy}{dx}=\frac{y'(t)}{x'(t)}=-\frac{3 \sin ^2\theta \cos \theta }{3 \cos ^2\theta \sin \theta}=-\frac{\sin \theta}{\cos \theta}=-\tan \theta$.
At $\theta =\frac{\pi }{3}$, the slope of the tangent line is $\frac{dy}{dx}|_{\theta =\frac{\pi }{3}}=-\sqrt{3}$.
The corresponding point is $c(\frac{\pi}{3})=\left(\frac{1}{8},\frac{3 \sqrt{3}}{8}\right)$.
So, the equation of the tangent line at $\theta =\frac{\pi }{3}$ is
$y-\frac{3 \sqrt{3}}{8}=-\sqrt{3} \left(x-\frac{1}{8}\right)$.
Or
$y=-\sqrt{3} x+\frac{\sqrt{3}}{2}$.
