For integer solutions:
You can rewrite the equations using
$$y^3 - 9x^2 +27x - 27 = y^3 - x^3 + (x-3)^3$$
and similarly for the two other equations. Adding the three equations then yields
$$(x-3)^3 + (y-3)^3 + (z-3)^3 = 0.$$
It is long known that in all integer solutions of the equation $a^3 + b^3 = c^3$, one of the integers must be $0$. Our equation is (modulo rearranging) exactly that, so we must have $x = 3$ or $y = 3$ or $z = 3$.
Suppose $x=3$. Then the first equation reduces to $y^3 - x^3 = 0$, hence $y^3 = 27$, hence $y = 3$. In the same way, it follows that $z = 3$. Starting from $y = 3$ or $z = 3$ yields the same. So $(3,3,3)$ is the only integer solution to the system.
Now, for any real solution, we still have by adding the equations
$$(x-3)^3 + (y-3)^3 + (z-3)^3 = 0,$$
so if one of $x,y,z$ were different from $3$, one of the three at least would be larger than $3$. Without loss of generality, let $x = 3 +\delta > 3$. Then, from the first equation, we obtain
y^3 &= x^3 - (x-3)^3\\
&= (3+\delta)^3 - \delta^3\\
&= 3^3 + 27\delta + 9\delta^2\\
so also $y > 3$. The same reasoning for the second equation then yields $z > 3$, and so
$$(x-3)^3 + (y-3)^3 + (z-3)^3 > 0$$
contradicting the assumption that $(x,y,z)$ is a solution of the system.
Hence $(3,3,3)$ is the only real solution of the system.