If $0 < x < \frac{\pi}{2}$ and $\frac{\sin x}{1 + \cos x} = \frac{1}{3}$, what is $\frac{\sin 2x}{1 + \cos 2x}$?
1. Given the equation:
   \[
   \frac{\sin x}{1 + \cos x} = \frac{1}{3}
   \]
   We recognize that:
   \[
   \frac{\sin x}{1 + \cos x} = \tan \frac{x}{2}
   \]
   Therefore:
   \[
   \tan \frac{x}{2} = \frac{1}{3}
   \]

2. We need to find:
   \[
   \frac{\sin 2x}{1 + \cos 2x}
   \]
   Using the double-angle identities for sine and cosine:
   \[
   \sin 2x = 2 \sin x \cos x
   \]
   \[
   \cos 2x = \cos^2 x - \sin^2 x
   \]
   However, we can use a simpler approach by recognizing another identity:
   \[
   \frac{\sin 2x}{1 + \cos 2x} = \tan x
   \]

3. We know that:
   \[
   \tan x = \frac{2 \tan \frac{x}{2}}{1 - \tan^2 \frac{x}{2}}
   \]
   Substituting \(\tan \frac{x}{2} = \frac{1}{3}\):
   \[
   \tan x = \frac{2 \cdot \frac{1}{3}}{1 - \left(\frac{1}{3}\right)^2}
   \]
   Simplify the expression:
   \[
   \tan x = \frac{\frac{2}{3}}{1 - \frac{1}{9}} = \frac{\frac{2}{3}}{\frac{8}{9}} = \frac{2}{3} \cdot \frac{9}{8} = \frac{2 \cdot 9}{3 \cdot 8} = \frac{18}{24} = \frac{3}{4}
   \]

4. Therefore:
   \[
   \frac{\sin 2x}{1 + \cos 2x} = \tan x = \frac{3}{4}
   \]

The final answer is \(\boxed{\frac{3}{4}}\)