What is the slope of the tangent line of r = - 2 sin (...

The slope of the tangent line of r at ${\displaystyle \theta =\frac{-\pi }{3}}$, is the same as the derivative of the function for that exact x-value. Therefore, we need to differentiate both sides of the function, so we get an expression for ${\displaystyle \frac{d}{d\theta }}$ r.
${\displaystyle \frac{d}{d\theta }\left[r\right]=\frac{d}{d\theta }[-2\sin \left(3\theta \right)-12\cos \left(\frac{\theta }{2}\right)]}$
${\displaystyle \frac{d}{d\theta }\left[r\right]=\frac{d}{d\theta }[-2\sin \left(3\theta \right)]-\frac{d}{d\theta }[-12\cos \left(\frac{\theta }{2}\right)]}$
From this point, we need to know how to differentiate ${\displaystyle \sin \theta }$ and ${\displaystyle \cos \theta }$:
${\displaystyle \frac{d}{d\theta }\sin \theta =\cos \theta }$
${\displaystyle \frac{d}{d\theta }\cos \theta =-\sin \theta }$
We need to know how the chain rule works as well. In this case, we have to substitute ${\displaystyle 3\theta }$ with u and ${\displaystyle \frac{\theta }{2}}$ with v. When we then differentiate, we have to also differentiate our substitute.
We then have:
${\displaystyle \frac{d}{d\theta }\left[r\right]=\frac{d}{d\theta }[-2\sin \left(u\right)]-\frac{d}{d\theta }\left[12\cos \left(v\right)\right]}$
${\displaystyle \frac{d}{d\theta }\left[r\right]=-2\cos \left(u\right)\cdot \frac{d}{d\theta }\left[u\right]-(-12\sin \left(v\right)\cdot \frac{d}{d\theta }\left[v\right])}$
Let's now substitute u and v back to our original functions.
${\displaystyle \frac{d}{d\theta }\left[r\right]=-2\cos \left(3\theta \right)\cdot \frac{d}{d\theta }\left[3\theta \right]-(-12\sin \left(\frac{\theta }{2}\right)\cdot \frac{d}{d\theta }\left[\frac{\theta }{2}\right])}$
${\displaystyle \frac{d}{d\theta }\left[r\right]=-2\cos \left(3\theta \right)\cdot 3+12\sin \left(\frac{\theta }{2}\right)\cdot \frac{1}{2}=-6\cos \left(3\theta \right)+6\sin \left(\frac{\theta }{2}\right)}$
Now we have an expression for ${\displaystyle \frac{d}{d\theta }r\left(\theta \right)}$, where we can put in any values we want, and get the slope for whatever ${\displaystyle \theta }$-value we want. Thus, let's put in ${\displaystyle \theta =\frac{-\pi }{3}}$:
${\displaystyle \frac{d}{d\theta }\left[r\left(\frac{-\pi }{3}\right)\right]}$
${\displaystyle =-6\cos (-\pi )+6\sin (-\frac{\pi }{6})}$
${\displaystyle \cos (-\pi )=-1}$ and ${\displaystyle \sin \left(\frac{-\pi }{6}\right)=-\frac{1}{2}}$
This gives us that the slope, for ${\displaystyle \theta =(-\frac{\pi }{3})}$, is:
${\displaystyle -6(-1)+6\cdot (-\frac{1}{2})=6-3=3}$