"The point [itex]2\pi[/itex]" doesn't make any sense. Was that a typo for [itex](2, \pi)[/itex], the given point? When [itex]\theta= \pi[/itex], [/itex]r= 2- 3cos(\pi)= 5[/itex] so, no, that point is not on the graph. However, [itex]r= 2- 3cos(\pi/2)= 2[/itex] so perhaps that was what was meant. Alternatively, perhaps the problem intended [itex]r= 2- 3sin(\theta)[/itex].
The slope of the tangent line at any point is, by definition, dy/dx. Since, in polar coordinates, [itex]x= r cos(\theta)[/itex] and [itex]y= r sin(\theta)[/itex], we have [itex]dx= dr cos(\theta)- r sin(\theta)d\theta[/itex] and [itex]dy= dr sin(\theta)d\theta+ r cos(\theta) d\theta[/itex] so that
[tex]\frac{dy}{dx}= \frac{cos(\theta) dr- r sin(\theta)d\theta}{sin(\theta)dr+ r cos(\theta)d\theta}= \frac{cos(\theta)\frac{dr}{d\theta}- r sin(\theta)}{sin(\theta)\frac{dr}{d\theta}+ r cos(\theta)}[/tex]
That can be simplified.