Solutions to $\frac1{\lfloor x\rfloor}+\frac1{\lfloor 2x\rfloor}=\{x\}+\frac13$

$$\dfrac{1}{\lfloor x\rfloor}+\dfrac{1}{\lfloor 2x\rfloor}=\{x\}+\dfrac{1}{3}\tag1$$

We have $\lfloor x\rfloor$ and $\lfloor 2x\rfloor$, so one way is to separate it into two cases :

Case 1 : $x=n+\alpha$ where $n\not=0\in\mathbb Z,0\le\alpha\lt 1/2$

Case 2 : $x=n+\alpha$ where $n\not=0\in\mathbb Z,1/2\le\alpha\lt 1$

For case 1, $$\begin{align}(1)&\implies \frac 1n+\frac{1}{2n}=\alpha+\frac 13\\&\implies \alpha=\frac{9-2n}{6n}\\&\implies0\le \frac{9-2n}{6n}\lt \frac 12\\&\implies (n,\alpha)=(2,5/12),(3,1/6),(4,1/24)\end{align}$$

I think that you can do for case 2 similarly.

We start by rewriting $\left\lfloor{x}\right\rfloor = k$ and $\{x\} = d$, with $k \in \Bbb{N}$ and $d \in [0, 1)$, where we exclude negative integers thanks to a comment on the OP. Now there are two cases: either $d \in [0, 0.5)$ or $d \in [0.5, 1)$. Notice that if $d \in [0.5, 1)$ then the integer part of $2x$ is not $2k$ but $2k + 1$.

Let us assume $d \in [0, 0.5)$ The equation then becomes

$$\frac1k + \frac1{2k} = d + \frac13 \iff \frac3{2k} = \frac{3d + 1}3 \iff$$

$$\iff \frac9{3d + 1} = 2k \iff k = \frac9{2(3d + 1)}$$

But given that $k$ is an integer, we must have that $6d + 2$ divides 9. Now you are left with very few possibilities that can be tested separately. Can you take it from here?

Also, do you think you can mimic this for the case $d \in [0.5, 1)$ making the necessary changes for the denominator of the second fraction?

EDIT As taking a similar path would make you go a looong way around, we try this other approach:

If $d \in [0.5, 1) $ then $\frac46 \leq d + \frac13 < \frac43$ therefore

$$\frac1k + \frac1{2k +1} \in [\frac46, \frac43) $$

Writing the two inequalities and multiplying by 6 one gets

$$4 \leq \frac6k + \frac6 {2k+1} < 8$$

One can easily spot that $k = 3$ is too big already thus you are left with a couple of values to try. If there is some integer solution for those inequalities, one then must find out how much $d $ would have to be. If after solving that $d \in [0.5, 1) $ you found another solution. Otherwise you did not.

Realize that RHS is positive so must be the LHS
So if $ \left \lfloor x\right \rfloor\geq 5$ then $$\frac{1}{\left \lfloor x\right \rfloor}+\frac{1}{\left \lfloor 2x\right \rfloor}< \frac{1}{3}$$ So then you just check the cases