In $\vartriangle ABC$, the external angle bisector of $\angle BAC$ intersects line $BC$ at $D$. $E$ is a point on ray $\overrightarrow{AC}$ such that $\angle BDE = 2\angle ADB$. If $AB = 10$, $AC = 12$, and $CE = 33$, compute $\frac{DB}{DE}$ .
1. **Construct Point \( F \)**: 
   - Construct point \( F \) on ray \(\overrightarrow{CA}\) such that \(\angle ADF = \angle ADB\). 
   - By construction, \(\triangle ADF \cong \triangle ADB\), so \(AF = AB = 10\) and \(DF = DB\).

2. **Calculate \( CF \)**:
   - Since \(CF = CA + AF\), we have \(CF = 12 + 10 = 22\).

3. **Use the Angle Bisector Theorem**:
   - The Angle Bisector Theorem states that the ratio of the segments created by the angle bisector on the opposite side is equal to the ratio of the other two sides of the triangle.
   - Applying this theorem, we get \(\frac{DF}{DE} = \frac{CF}{CE}\).

4. **Substitute Known Values**:
   - We know \(CF = 22\) and \(CE = 33\).
   - Therefore, \(\frac{DF}{DE} = \frac{22}{33} = \frac{2}{3}\).

5. **Conclusion**:
   - Since \(DF = DB\), we have \(\frac{DB}{DE} = \frac{2}{3}\).

The final answer is \(\boxed{\frac{2}{3}}\).