Solutions to $\frac1{\lfloor x\rfloor}+\frac1{\lfloor 2x\rfloor}=\{x\}+\frac13$ 
Find all solutions to $$\dfrac{1}{\lfloor x\rfloor}+\dfrac{1}{\lfloor 2x\rfloor}=\{x\}+\dfrac{1}{3}$$

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Unfortunately I have no idea as to how to go about this. On rearranging, I got $$3\lfloor 2x\rfloor = 3\lfloor x\rfloor\{x\}-2\lfloor x\rfloor$$
I'm not sure about what to do with the $3\lfloor 2x\rfloor $ term; I'd prefer to resolve it in terms of $\lfloor x\rfloor $ but am not able to. All that struck me was using the identity for $\lfloor nx\rfloor, n\in \Bbb Z$. However on first glance, it did not strike me as particularly useful.$$$$
I would be grateful for any help. Many thanks! 

I decided to do @mathlove's case $2$.
CASE $2$
We see that $n \gt 0$.
$$\dfrac{1}{n} + \dfrac{1}{2n + 1}= \delta + \frac 13$$
$$\dfrac{3n+1}{2n^2+n} - \dfrac 13  = \delta$$
$$ \delta = \frac{-2n^2 + 8n + 3}{6n^2 + 3n}$$
\begin{align}
   0 \le \frac{-2n^2 + 8n + 3}{6n^2 + 3n} \lt 1 \\
   0 \le -2n^2 + 8n + 3 \lt 6n^2 + 3n \\
\end{align}
\begin{align}
   0 &\le -2n^2 + 8n + 3 \\
   2n^2 - 8n - 3 &\le 0 \\
   n^2 - 4n &\le \frac 32 \\
   (n - 2)^2 &\le \frac{11}{2} \\
   (n - 2)^2 &\le 4 \\
   n &\in \{1,2,3,4\}
\end{align}
\begin{align}
   -2n^2 + 8n + 3 &\lt 6n^2 + 3n \\
   8n^2 - 5n - 3 &\gt 0 \\
   256n^2 - 160n &\gt 96 \\
   (16n - 5)^2 &\gt 96 \\
   16n - 5 &\gt 10 \\
   16n &\gt 15 \\
   n &\ge 1 
\end{align}
$n= -1 \implies \delta = -\frac 73$
$n=1 \implies \delta = 1$
$n=2 \implies \delta = \frac{11}{30} \implies x = 2\frac{11}{30}$
$n=3 \implies \delta = \frac 17 \implies x = 3\frac 17$
$n=4 \implies \delta = \frac{1}{36} \implies x = 1\frac{1}{36}$
$n=5 \implies \delta = - \frac{7}{165}$
So $n \in \{ 2\frac{11}{30}, 3\frac 17,  1\frac{1}{36}\}$
