Find the equation (in terms of xx and yy) of the tangent line to the curve r=3sin2θr=3sin⁡2θ at θ=π/6θ=π/6

The slope of the tangent line to a curve is the first derivative, dy/dx, of the curve at the point of tangency. 

In this problem the point of tangency has theta = pi/6, so the corresponding r = 5 sin[(3)(pi/6)] = 5. 
Converting these (r, theta) coordinates to (x, y) coordinates you use the polar to cartesian coordinate conversion eqns.: 
x = r cos(theta) and y = r sin(theta).  
So x = 5 cos(pi/6) = 4.33; y = 5 sin(pi/6) = 2.5 
So the point of tangency where the tangent line intersects the curve is (x,y) = (4.33, 2,5) 

Using the formula from the attached website, which I assume your teacher derived in class: 
dy/dx = [[dr/d(theta)] sin(theta) + r cos(theta)] / [[dr/d(theta)] cos(theta) - r sin(theta)]  
From r = 5sin(3 theta) , dr/d(theta) = 15 cos(3 theta) 
dy/dx = [ [15 cos(3 theta)] sin(theta) + r cos(theta)] / [ [15 cos(3 theta)] cos(theta) - r sin(theta)]  
dy/dx evaluated at theta = pi/6 and r = 5 is:  
dy/dx = [ [15 cos(3 pi/6] sin(pi/6) + 5 cos(pi/6)] / [ [15 cos(3 pi/6)] cos(pi/6) - 5 sin(pi/6)]  
dy/dx = - cot(pi/6) = -1.732 

So we have the tangent line of the form y = (-1.732)x + b where the point (x,y) = (4.33, 2.5) is on the line. 
b = 2.5 + (1.732)(4.33) = 10 

So the tangent line is y = -1.732 x + 10