Answer
The slope of the tangent line is $~\frac{\sqrt{3}}{9}$
Work Step by Step
$r = 1+2~cos~\theta$
$x = r~cos~\theta = (1+2~cos~\theta)~(cos~\theta)$
$y = r~sin~\theta = (1+2~cos~\theta)~(sin~\theta)$
We can find $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-2~sin^2~\theta+cos~\theta+2~cos^2~\theta}{-2~sin~\theta~cos~\theta-sin~\theta-2~sin~\theta~cos~\theta} = \frac{2~sin^2~\theta-cos~\theta-2~cos^2~\theta}{4~sin~\theta~cos~\theta+sin~\theta}$
When $\theta = \frac{\pi}{3}$:
$\frac{dy}{dx} =\frac{2~sin^2~\frac{\pi}{3}-cos~\frac{\pi}{3}-2~cos^2~\frac{\pi}{3}}{4~sin~\frac{\pi}{3}~cos~\frac{\pi}{3}+sin~\frac{\pi}{3}}$
$\frac{dy}{dx} =\frac{2~(\frac{\sqrt{3}}{2})^2-\frac{1}{2}-2~(\frac{1}{2})^2}{4~(\frac{\sqrt{3}}{2})~(\frac{1}{2})+(\frac{\sqrt{3}}{2})}$
$\frac{dy}{dx} =\frac{\frac{3}{2}-\frac{1}{2}-\frac{1}{2}}{\frac{3\sqrt{3}}{2}}$
$\frac{dy}{dx} =\frac{1}{3\sqrt{3}}$
$\frac{dy}{dx} =\frac{1}{3\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}$
$\frac{dy}{dx} =\frac{\sqrt{3}}{9}$
The slope of the tangent line is $~\frac{\sqrt{3}}{9}$
