Answer
The slope of the tangent line is $~-\sqrt{3}$
Work Step by Step
$r = cos~\frac{\theta}{3}$
$x = r~cos~\theta = cos~\frac{\theta}{3}~cos~\theta$
$y = r~sin~\theta = cos~\frac{\theta}{3}~sin~\theta$
We can find $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-\frac{1}{3}~sin~\frac{\theta}{3}~sin~\theta+cos~\frac{\theta}{3}~cos~\theta}{-\frac{1}{3}~sin~\frac{\theta}{3}~cos~\theta-cos~\frac{\theta}{3}~sin~\theta}$
When $\theta = \pi$:
$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-\frac{1}{3}~sin~\frac{\pi}{3}~sin~\pi+cos~\frac{\pi}{3}~cos~\pi}{-\frac{1}{3}~sin~\frac{\pi}{3}~cos~\pi-cos~\frac{\pi}{3}~sin~\pi}$
$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-\frac{1}{3}~(\frac{\sqrt{3}}{2})~(0)+( \frac{1}{2})~(-1)}{-\frac{1}{3}~(\frac{\sqrt{3}}{2})~(-1)-( \frac{1}{2})~(0)}$
$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-\frac{1}{2}}{\frac{\sqrt{3}}{6}}$
$\frac{dy}{dx} =\frac{-3}{\sqrt{3}}$
$\frac{dy}{dx} =-\sqrt{3}$
The slope of the tangent line is $~-\sqrt{3}$
