Let $x, y$ and $z$ be three real numbers such that $x^{2}+y^{2}+z^{2}=1$. Find the minimum and maximum possible values of the real number $x y+y z-z x$.
Based on the identity

$$
(x-y+z)^{2}=x^{2}+y^{2}+z^{2}-2(x y+y z-z x)=1-2(x y+y z-z x),
$$

we are to find the extreme values that the number $(x-y+z)^{2}$ can take.
First, since $(x-y+z)^{2} \geqslant 0$, with equality when $x=y=1 / \sqrt{2}$ and $z=0$, we have

$$
x y+y z-z x=\frac{1-(x-y+z)^{2}}{2} \leqslant \frac{1}{2}
$$

with equality when $x=y=1 / \sqrt{2}$ and $z=0$.
Furthermore, the Cauchy-Schwarz inequality indicates that

$$
(x-y+z)^{2} \leqslant\left(x^{2}+y^{2}+z^{2}\right) \times\left(1^{2}+(-1)^{2}+1^{2}\right)=3,
$$

with equality if and only if the triplets $(x, y, z)$ and $(1,-1,1)$ are collinear. Therefore, we have

$$
(x y+y z-z x)=\frac{1-(x-y+z)^{2}}{2} \geqslant \frac{1-3}{2}=-1
$$

with equality when $x=-y=z=1 / \sqrt{3}$.
In conclusion, the sought extreme values are -1 and $1 / 2$.