• Divide the problem into two cases based on the difference of $\lfloor x \rfloor$ and $\left\lfloor x+\frac{1}{3}\right\rfloor$.
• Solve the equations using divisibility and wrap up the solutions for $x$.

The answer sets are these three intervals: $ \left[ 1\dfrac{2}{3}, 2 \right), \left[ 2\dfrac{2}{3}, 3 \right)\text{ and }\left[ -2\dfrac{1}{3}, -2 \right)$. 

We prove that these are the only. Let $x=a+b$ where $a=\lfloor x \rfloor$ and $b=\{x\}=x-\lfloor x \rfloor$. There are two cases.

1.   $b \geq \frac{2}{3}$. In this case, $\left\lfloor x+\frac{1}{3}\right\rfloor = a+1$. So the equation becomes

             $a^3 - 7(a+1) = -13 $

             $\implies a^3 -7a = -6 $

            $a$ is an integer and the left side is divisible by $a$, so the right side must be too. Checking the divisors of $-6$, we find $a=1, 2, -3$ are the possible solutions for $a$. These lead to the answers we mentioned before.

2.   $b < \frac{2}{3}$. Here, $\left\lfloor x+\frac{1}{3}\right\rfloor = a$. So \[a^3-7a = -13.\] After checking the divisors of $-13$, we find that this case leads to no solutions.