Find the slope of the tangent line to the polar curve for the given value of $\displaystyle \theta$
$\displaystyle r=1/\theta; \theta=2$
$\displaystyle \frac{dy}{dx}=\frac{r'(\theta)\sin\theta+r(\theta) \cos\theta}{r'(\theta)\cos\theta-r(\theta)\sin\theta} $
So in our case $\displaystyle \frac{dy}{dx}=\frac{-\tfrac{1}{\theta^2}\sin\theta+\tfrac{1}{\theta}\co s\theta}{-\tfrac{1}{\theta^2}\cos\theta-\tfrac{1}{\theta}\sin\theta} $
So just plug in $\displaystyle \theta=2 $ to get your answer.