Let $S=x+y+z$. We add to get $\frac{1}{3}S^2\leqslant x^2+y^2+z^2\leqslant 2S$ and thus $S\leqslant 6$ and $z^2\leqslant S-z\leqslant 6-z$ which means $z\leqslant 2$ and $(2,2,2)$ satisfies all conditions, hence $z=2$ is the maximum. For minimum we have $2z\geqslant x^2-y+y^2-x\geqslant -\frac{1}{4}-\frac{1}{4}$. We can easily see that $ (\frac{1}{2},\frac{1}{2},-\frac{1}{4}) $ satistfies all conditions and so minimum is $z=-\frac{1}{4}$.