In a tetrahedral $ABCD$, given that $\angle ADB=\angle BDC =\angle CDA=\frac{\pi}{3}$, $AD=BD=3$, and $CD=2$. Find the radius of the circumsphere of $ABCD$.
1. **Given Data and Initial Calculations:**
   - We are given a tetrahedron \(ABCD\) with \(\angle ADB = \angle BDC = \angle CDA = \frac{\pi}{3}\).
   - The side lengths are \(AD = BD = 3\) and \(CD = 2\).
   - Using the cosine theorem, we can find the other side lengths:
     \[
     AD^2 = BD^2 = 9, \quad CD^2 = 4
     \]
     \[
     \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}
     \]
     \[
     AB^2 = AD^2 + BD^2 - 2 \cdot AD \cdot BD \cdot \cos\left(\frac{\pi}{3}\right) = 9 + 9 - 2 \cdot 3 \cdot 3 \cdot \frac{1}{2} = 9
     \]
     \[
     AC^2 = AD^2 + CD^2 - 2 \cdot AD \cdot CD \cdot \cos\left(\frac{\pi}{3}\right) = 9 + 4 - 2 \cdot 3 \cdot 2 \cdot \frac{1}{2} = 7
     \]
     \[
     BC^2 = BD^2 + CD^2 - 2 \cdot BD \cdot CD \cdot \cos\left(\frac{\pi}{3}\right) = 9 + 4 - 2 \cdot 3 \cdot 2 \cdot \frac{1}{2} = 7
     \]

2. **Cayley-Menger Determinant:**
   - The circumsphere of tetrahedron \(ABCD\) is denoted as \((O, R)\).
   - The Cayley-Menger determinant for a tetrahedron is used to find the radius \(R\) of the circumsphere.
   - The determinant is given by:
     \[
     V_{4}^{2} \sim \left|\begin{array}{cccccc}
     0 & 1 & 1 & 1 & 1 & 1 \\
     1 & 0 & AB^2 & AC^2 & AD^2 & AO^2 \\
     1 & AB^2 & 0 & BC^2 & BD^2 & BO^2 \\
     1 & AC^2 & BC^2 & 0 & CD^2 & CO^2 \\
     1 & AD^2 & BD^2 & CD^2 & 0 & DO^2 \\
     1 & AO^2 & BO^2 & CO^2 & DO^2 & 0
     \end{array}\right|
     \]
   - Substituting the known values:
     \[
     \left|\begin{array}{cccccc}
     0 & 1 & 1 & 1 & 1 & 1 \\
     1 & 0 & 9 & 7 & 9 & R^2 \\
     1 & 9 & 0 & 7 & 9 & R^2 \\
     1 & 7 & 7 & 0 & 4 & R^2 \\
     1 & 9 & 9 & 4 & 0 & R^2 \\
     1 & R^2 & R^2 & R^2 & R^2 & 0
     \end{array}\right| = 0
     \]

3. **Simplifying the Determinant:**
   - By adding/subtracting rows or columns to simplify the determinant, we aim to get at most one non-zero element in a row or column.
   - After simplification, we find:
     \[
     -2^5 \cdot 3^4 \cdot (R^2 - 3) = 0
     \]
   - Solving for \(R\):
     \[
     R^2 - 3 = 0 \implies R^2 = 3 \implies R = \sqrt{3}
     \]

The final answer is \(\boxed{\sqrt{3}}\)