One, (20 points) Let $x, y$ be real numbers, and $x^{2}+x y+y^{2}=1$. Find the range of values for $x^{2}-x y+y^{2}$.
Given $x^{2}+x y+y^{2}=1$.
Let $x^{2}-x y+y^{2}=\hbar$.
(1) - (2), rearranging gives $x y=\frac{1-k}{2}$.

From (1), we have $(x+y)^{2}=x y+1$.
Substituting (3) into (1), we get
$$
(x+y)^{2}=\frac{3-k}{2} .
$$

Since $x, y$ are real numbers, $(x+y)^{2} \geqslant 0$.
From (5), we get $k \leqslant 3$.
Under the condition (8), from (5) we get
$$
x+y= \pm \frac{\sqrt{3-k}}{\sqrt{2}} .
$$

From (3) and (7), $x, y$ are the roots of the quadratic equation
$$
t^{2} \mp \frac{\sqrt{3-k}}{\sqrt{2}} t+\frac{1-k}{2}=0
$$

Since $t$ is a real number, the discriminant is
$$
\Delta=\frac{3-k}{2}-2(1-k) \geqslant 0 .
$$

Thus, $k \geqslant \frac{1}{3}$.
From ( ) and (8), we have $-\frac{1}{3} \leqslant k \leqslant 3$,
Therefore, $-\frac{1}{3} \leqslant x^{2}-x y+y^{2} \leqslant 3$.