Answer
$-\frac{\sqrt {2}}{2}$
Work Step by Step
Step 1. Letting $cos^{-1}(-\frac{\sqrt 3}{3})=t$, we have $cos(t)=-\frac{\sqrt 3}{3}$ and $\frac{\pi}{2}\lt t\lt\pi$.
Step 2. Let $x=\sqrt 3, r=3$, and $y=\sqrt {r^2-x^2}=\sqrt {6}$.
Step 3. Sides $x,y,r$ form a right triangle with angle $|t|$ facing $y$.
Step 4. We know $cot(t)$ is negative and we have $cot(t)=-\frac{x}{y}=-\frac{\sqrt {2}}{2}$ or $cot(cos^{-1}(-\frac{\sqrt 3}{3}))=-\frac{\sqrt {2}}{2}$
