Find all the real numbers $x$ such that $\frac{1}{[x]}+\frac{1}{[2x]}=\{x\}+\frac{1}{3}$ where $[x]$ denotes the integer part of $x$ and $\{x\}=x-[x]$.
For example, $[2.5]=2, \{2.5\} = 0.5$ and $[-1.7]= -2, \{-1.7\} = 0.3$
To solve the equation \(\frac{1}{[x]} + \frac{1}{[2x]} = \{x\} + \frac{1}{3}\), we need to consider the properties of the floor function \([x]\) and the fractional part function \(\{x\}\).

1. **Case Analysis Based on the Value of \(x\)**:
   - If \(x < 0\), the left-hand side of the equation is negative while the right-hand side is positive, leading to a contradiction. Therefore, \(x \geq 0\).
   - If \(0 \leq x < 1\), then \([x] = 0\), making the left-hand side undefined. Thus, \(x \geq 1\).

2. **Considering \(x \geq 1\)**:
   - If \(1 \leq x < 2\), then \([x] = 1\) and \([2x] = 2\). The left-hand side is \(\frac{1}{1} + \frac{1}{2} = 1.5\), but the right-hand side \(\{x\} + \frac{1}{3}\) is less than \(1.333\) since \(\{x\} < 1\). Thus, no solutions exist in this range.
   - Therefore, \(x \geq 2\).

3. **Simplifying the Equation for \(x \geq 2\)**:
   - The left-hand side is at most \(\frac{1}{2} + \frac{1}{4} = 0.75\). Thus, \(\{x\} \leq 0.75 - \frac{1}{3} = \frac{5}{12}\).
   - This implies \([2x] = 2[x]\) because \(\{x\} < 0.5\).

4. **Solving for Specific Values of \([x]\)**:
   - Let \([x] = k\). Then \([2x] = 2k\) and the equation becomes:
     \[
     \frac{1}{k} + \frac{1}{2k} = \{x\} + \frac{1}{3}
     \]
     Simplifying, we get:
     \[
     \frac{3}{2k} = \{x\} + \frac{1}{3}
     \]
     \[
     \{x\} = \frac{3}{2k} - \frac{1}{3}
     \]

5. **Checking for Integer Values of \(k\)**:
   - For \(k = 2\):
     \[
     \{x\} = \frac{3}{4} - \frac{1}{3} = \frac{9}{12} - \frac{4}{12} = \frac{5}{12}
     \]
     Thus, \(x = 2 + \frac{5}{12} = \frac{29}{12}\).

   - For \(k = 3\):
     \[
     \{x\} = \frac{3}{6} - \frac{1}{3} = \frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}
     \]
     Thus, \(x = 3 + \frac{1}{6} = \frac{19}{6}\).

   - For \(k = 4\):
     \[
     \{x\} = \frac{3}{8} - \frac{1}{3} = \frac{9}{24} - \frac{8}{24} = \frac{1}{24}
     \]
     Thus, \(x = 4 + \frac{1}{24} = \frac{97}{24}\).

   - For \(k \geq 5\):
     \[
     \{x\} = \frac{3}{2k} - \frac{1}{3} \leq \frac{3}{10} - \frac{1}{3} = \frac{9}{30} - \frac{10}{30} = -\frac{1}{30} < 0
     \]
     This is a contradiction since \(\{x\}\) must be non-negative.

Thus, the solutions are \(x = \frac{29}{12}\), \(x = \frac{19}{6}\), and \(x = \frac{97}{24}\).

The final answer is \(\boxed{\left\{\frac{29}{12}, \frac{19}{6}, \frac{97}{24}\right\}}\).