To solve this problem, we need to calculate the curvature $\kappa(t)$ of the given curve $\mathbf{r}(t) = \langle 3\cos(t), 2\sin(t) \rangle$ at $t = \frac{\pi}{3}$. The curvature formula for a plane curve parameterized by $\mathbf{r}(t) = \langle x(t), y(t) \rangle$ is:

$\kappa(t) = \frac{\|\mathbf{r}'(t) \times \mathbf{r}''(t)\|}{\|\mathbf{r}'(t)\|^3}.$

Step 1: Compute $\mathbf{r}'(t)$

The first derivative of $\mathbf{r}(t)$ is: $\mathbf{r}'(t) = \frac{d}{dt}\langle 3\cos(t), 2\sin(t) \rangle = \langle -3\sin(t), 2\cos(t) \rangle.$

Step 2: Compute $\mathbf{r}''(t)$

The second derivative of $\mathbf{r}(t)$ is: $\mathbf{r}''(t) = \frac{d}{dt}\langle -3\sin(t), 2\cos(t) \rangle = \langle -3\cos(t), -2\sin(t) \rangle.$

Step 3: Compute $\|\mathbf{r}'(t)\|$

The magnitude of $\mathbf{r}'(t)$ is: $\|\mathbf{r}'(t)\| = \sqrt{(-3\sin(t))^2 + (2\cos(t))^2} = \sqrt{9\sin^2(t) + 4\cos^2(t)}.$

At $t = \frac{\pi}{3}$: $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}, \quad \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}.$ Substitute these values: $\|\mathbf{r}'\left(\frac{\pi}{3}\right)\| = \sqrt{9\left(\frac{\sqrt{3}}{2}\right)^2 + 4\left(\frac{1}{2}\right)^2} = \sqrt{9\frac{3}{4} + 4\frac{1}{4}} = \sqrt{\frac{27}{4} + \frac{4}{4}} = \sqrt{\frac{31}{4}} = \frac{\sqrt{31}}{2}.$

Step 4: Compute $\|\mathbf{r}'(t) \times \mathbf{r}''(t)\|$

The cross product in 2D can be expressed as the determinant: