Let $ x\in\mathbb{R} . $ Then, there exists $ k\in\mathbb{Z} $ such that $ x\in\left[\frac{k}{2},\frac{k+1}{2}\right). $ Case I: $ \exists l\in\mathbb{Z}\quad k=2l. $ $ \{ x\} =\frac{1}{l} +\frac{1}{2l}-\frac{1}{3}=\frac{2}{l}-1\in [0,1)\implies l\in (1,2]\cap\mathbb{Z}=\{ 2\} . $ So, $ x\in [2,3) $ and $ \{ x\}=\frac{1}{2} +\frac{1}{4} -\frac{1}{3}=\frac{5}{12}\implies x=2+\frac{5}{12}=\frac{29}{12} .\quad\text{(*)} $ Case II: $ \exists l\in\mathbb{Z}\quad k=2l+1. $ $ \{ x\}=\frac{1}{l}+\frac{1}{2l+1}-\frac{1}{3}=\frac{-2l^2+8l+3}{6l^2+3l}\in [0,1) . $ A study of thee roots of the numerator and denominator (which I'll omit), shows us that $ l\in\{-3,-2,-1,1,2,3\} . $ Verifying all of those diligently (which, again, if you're interested, you can do it for yourself), we found that none give solutions. So, solution $ \text{(*)} $ is unique.

CatalinBordea wrote: $\ldots$ Case I: $ \exists l\in\mathbb{Z}\quad k=2l. $ $ \{ x\} =\frac{1}{l} +\frac{1}{2l}-\frac{1}{3}=\frac{2}{l}-1\in [0,1)\implies l\in (1,2]\cap\mathbb{Z}=\{ 2\} . $ So, $ x\in [2,3) $ and $ \{ x\}=\frac{1}{2} +\frac{1}{4} -\frac{1}{3}=\frac{5}{12}\implies x=2+\frac{5}{12}=\frac{29}{12} .\quad\text{(*)} $ Case II: $ \exists l\in\mathbb{Z}\quad k=2l+1. $ $\ldots$ CatalinBordea is missing some solutions. I suppose they were brought on by the error in the first equation under Case I. Here is my solution: If $x<0$, then the left hand-side is negative while the right hand side is positive, so no solutions. If $0\le x<1$, then the left hand-side is undefined. If $1\le x<2$, then the left hand-side is at least $\frac{1}{1}+\frac{1}{3}$, but the right hand-side cannot reach this, since $\{x\}<1$. Thus $x\ge 2$. Now, the left hand-side is at most $\frac{1}{2}+\frac{1}{4}$, so $\{x\}\le \frac{1}{2}+\frac{1}{4}-\frac{1}{3}=\frac{5}{12}<\frac{1}{2}$. This means that $[2x]=2[x]$ and the left hand-side simplifies to $\frac{3}{2[x]}$. If $[x]=2$, then $\{x\}=\frac{3}{4}-\frac{1}{3}=\frac{5}{12}$. If $[x]=3$, then $\{x\}=\frac{3}{6}-\frac{1}{3}=\frac{1}{6}$. If $[x]=4$, then $\{x\}=\frac{3}{8}-\frac{1}{3}=\frac{1}{24}$. If $[x]\ge 5$, then $\{x\}\le \frac{3}{10}-\frac{1}{3}<0$, which is a contradiction. Thus the solutions are $2\frac{5}{12}$, $3\frac{1}{6}$ and $4\frac{1}{24}$.

Quote: Thus the solutions are $2\frac{5}{12}$, $3\frac{1}{6}$ and $4\frac{1}{24}$. You mean $ 2+\frac{5}{12}, $ etc I was almost sure that I had an error, since that was calculated quickly.