The equations $r = 3\sin(2\theta)$ and $\frac{d\theta}{dt} = 2$ describe the motion of a particle in polar coordinates. Find the velocity of the particle in terms of the unit vectors $u_r$ and $u_\theta$. Sketch the path of the particle.
I don't understand how to relate the unit vectors to the position function or how to use $\frac{d\theta}{dt}. $
In polar coordinates, the radial unit vector is $$\vec u_r(\theta)=\begin{pmatrix}\cos\theta\\\sin\theta\end{pmatrix}\tag{1}.$$ Therefore a point located at radius $r$ and angle $\theta$ is located at the point of coordinates $\vec r=\begin{pmatrix}r\cos\theta\\r\sin\theta\end{pmatrix}$. We have the relation $\boxed{\vec r=r\vec u_r(\theta)}$.
The motion of the particle is described by $r(\theta)=3\sin(2\theta)$ that is as function of time $r(t)=3\sin(2\theta(t)).$ The position is therefore $$\vec r(t)=r(t)\,\vec u_r(\theta(t)).$$ Now take the derivative of this expression using the definition (1) and note that the derivative of $\vec u_r$ is $$ \frac{\mathrm d\vec u_r}{\mathrm d\theta}(\theta)=\begin{pmatrix}-\sin\theta\\\cos\theta\end{pmatrix}=\vec u_\theta(\theta).$$ This last formula will enter into play in $\frac{\mathrm d\vec u_r(\theta(t))}{\mathrm dt}=\frac{\mathrm d\theta(t)}{\mathrm dt}\;\vec u_\theta(\theta(t))$.