If $x, y$ are real numbers, and $1 \leqslant x^{2}+4 y^{2} \leqslant 2$. Find the range of $x^{2}-2 x y+4 y^{2}$.
Let $x^{2}+4 y^{2}=m$,
$$
x^{2}-2 x y+4 y^{2}=n.
$$
(1)-(2), we get $x \cdot 2 y=m-n$.
From (1),
$$
(x+2 y)^{2}=m+4 x y=3 m-2 n,
$$

thus $x+2 y= \pm \sqrt{3 m-2 n}$.
Therefore, $x, 2 y$ are the two real roots of the equation $t^{2} \pm \sqrt{3 m-2 n} t+(m-n)=0$.
By $\Delta=3 m-2 n-4(m-n) \geqslant 0$ and $3 m-2 n \geqslant 0$, we get $\frac{1}{2} m \leqslant n \leqslant \frac{3}{2} m$.
Given $1 \leqslant m \leqslant 2$, then $\frac{1}{2} \leqslant n \leqslant 3$,
hence the range of $x^{2}-2 x y+4 y^{2}$ is $\left[\frac{1}{2}, 3\right]$.