Disclaimer: The solutions we've shared are just one exciting approach, and there are surely many other wonderful methods out there. We’d love to hear your alternative solutions in the community thread below, so let's keep the creativity flowing!

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.