Answer
$$y=-\sqrt 3x +\frac{ \sqrt 3}{2}$$
Work Step by Step
Recall, to find the equation of a line, we need to find the slope of the line and a point that lies on the line.
Now, we want to find the equation of the line tangent to the curve at the point where $\theta =\frac{\pi }{6}$.
So, we know the slope of this line is the derivative of the curve at the point where $\theta =\frac{\pi }{6}$, and because this line is tangent to the curve at the point where $\theta =\frac{\pi }{6}$, we know the point where $\theta =\frac{\pi }{6}$ lies on this line.
So, let us first find the derivative at $\theta =\frac{\pi }{6}$. The curve is given by the parametric equations $$x=\sin ^3 \theta$$ and $$y=\cos ^3 \theta.$$ So, the derivative at $\theta =\frac{\pi }{6}$ is $$\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}},$$ evaluated at $\theta =\frac{\pi }{6}$
So, we find the derivatives of each of the parametric equations with respect to $\theta$. $$\frac{dx}{d\theta}=\frac{d}{d\theta}(\sin ^3 \theta )=3\sin ^2 \theta \frac{d}{d\theta}(\sin \theta)=3\sin^2 \theta \cos \theta$$ and $$\frac{dy}{d\theta}=\frac{d}{d\theta}(\cos ^3 \theta )=3\cos ^2 \theta \frac{d}{d\theta}(\cos \theta)=3\cos^2 \theta (-\sin \theta)$$ $$=-3\cos ^2 \theta \sin \theta.$$ Then $$\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}=\frac{-3\cos ^2 \theta \sin \theta}{3\sin^2 \theta \cos \theta}.$$ Next we plug in $\theta =\frac{\pi }{6}$ to get $$\frac{-3(\cos ^2 \frac{\pi }{6}) \sin \frac{\pi }{6}}{3(\sin^2 \frac{\pi }{6}) \cos \frac{\pi }{6}}=\frac{-3(\frac{\sqrt 3}{2})^2 (\frac{1}{2})}{3(\frac{1}{2})^2 (\frac{\sqrt 3}{2})}=\frac{-3(\frac{3}{4}) (\frac{1}{2})}{3(\frac{1}{4}) (\frac{\sqrt 3}{2})}.$$ We cancel the $3's$ and multiply the top and bottom by $8$ to get $$\frac{-(3)(1)}{(1)(\sqrt 3)}=\frac{-3}{\sqrt 3}.$$ We multiply the top and bottom by $\sqrt 3$ to get $-\sqrt 3$. Thus, the slope of our line is $-\sqrt 3$.
Now, to find the point that lies on the line, we plug $\theta =\frac{\pi }{6}$ into our parametric equations. We get $$x=\sin ^3 \theta=\sin ^3 (\frac{\pi }{6})=(\frac{1}{2})^3=\frac{1}{8}$$ and $$y=\cos ^3 \theta=\cos ^3 (\frac{\pi }{6})=(\frac{\sqrt 3}{2})^3=\frac{3 \sqrt 3}{8}.$$ Thus, our point is $(\frac{1}{8} , \frac{3 \sqrt 3}{8})$.
Finally, we are ready to find the equation of our line. We will use the formula $$y-y_{1}=m(x-x_{1}),$$ where $m$ is the slope and $(x_{1},y_{1})$ is the point that lies on the line.
For us, $m=-\sqrt 3$ and $(x_{1},y_{1})=( \frac{1}{8}, \frac{3 \sqrt 3}{8})$. We plug these into the equation to get $$y-\frac{3 \sqrt 3}{8}=-\sqrt 3 (x- \frac{1}{8}).$$ Distributing the $-\sqrt 3$ on the right, we get $$y-\frac{3 \sqrt 3}{8}=-\sqrt 3 x +\frac{\sqrt 3}{8}.$$ Adding $\frac{3 \sqrt 3}{8}$ to both sides, we get $$y=-\sqrt 3x +\frac{4 \sqrt 3}{8}\\\\=y=-\sqrt{3x}+\frac{\sqrt{3}}{2}.$$
