Assume that $x=\max(x,y,z)$ and $z=\min(x,y,z)$. The other cases can be solved analogously. Note that: $$ x^3+x^2+x \ge x+y^2+z^3 =3 \implies x^3+x^2+x-3=(x-1)((x+1)^2+1) \ge 0 $$This means that $x \ge 1$. Similarly: $$ z+z^2+z^3 \le x+y^2+z^3 =3 \implies (z-1)((z+1)^2+1) \le 0 $$This means that $z \le 1$. Consider the case, when $y \le 1$. Then subtracting the second equation from the first gives us: \begin{align*} z^3-z^2 +y^2-y=x^3-x^2 \\ z^2(z-1)+y(y-1) =x^2(x-1) \end{align*}Since $z \le 1$, $y \le 1$ and $x \ge 1$, then the LHS of the the last expression is $\le 0$, whereas the RHS is $\ge 0$. The equality of both sides is possible if and only if $x=y=z=1$. Consider the case when $y \ge 1$. Then subtracting the first equation from the third gives us: \begin{align*} x^2-x +y^3-y^2 = z^3-z \\ x(x-1)+y^2(y-1)=z(z+1)(z-1) \end{align*}Since $x \ge 1$, $y \ge 1$ and $z \le 1$, then the LHS of the last expression is $\ge 0$, whereas the RHS is $\le 0$. The equality of both sides is possible if and only if $x=y=z=1$. We conclude that the only triplet satisfying the given system of equations is $(x,y,z)=(1,1,1)$.