Three, (50 points) Let $a, b$ be two given positive real numbers, and real numbers $x, y$ satisfy $a x^{2}-b x y+a y^{2}=1$. Try to find the range (value range) of $f=x^{2}+y^{2}$.
Three, Solution: Let $x=\frac{m+n}{\sqrt{a}}, y=\frac{m-n}{\sqrt{a}}$, then from the given equation we get: $a \cdot \frac{(m+n)^{2}}{a}-b \cdot \frac{m^{2}-n^{2}}{a}+a \cdot \frac{(m-n)^{2}}{a}=1$,
simplifying to $(2 a-b) \cdot m^{2}+(2 a+b) \cdot n^{2}=a$. (1)
At this point, $f=x^{2}+y^{2}=\frac{(m+n)^{2}+(m-n)^{2}}{a}=\frac{2}{a} \cdot\left(m^{2}+n^{2}\right)$.
(1) If $2 a-b>0$, i.e., $b<2a$, then from (1) we have $\frac{a}{2 a+b} \leqslant m^{2}+n^{2} \leqslant \frac{a}{2 a-b}$, (3)
thus $\frac{2}{2 a+b} \leqslant f=\frac{2}{a} \cdot\left(m^{2}+n^{2}\right) \leqslant \frac{2}{2 a-b}$, so the range of $f$ is $\left[\frac{2}{2 a+b}, \frac{2}{2 a-b}\right]$.
(2) If $2 a-b \leqslant 0$, i.e., $b \geqslant 2 a$, then from (1) we get:
$$
(2 a+b) n^{2}=a+(b-2 a) \cdot m^{2} \geqslant a \text {, (4) }
$$

i.e., $n^{2} \geqslant \frac{a}{2 a+b}$, (5)
from (4) we know: $m^{2}$ can take any non-negative real number $(\because b-2 a \geqslant 0)$,
so $f=\frac{2}{a} \cdot\left(m^{2}+n^{2}\right)=\frac{2}{a} \cdot\left[m^{2}+\frac{a+(b-2 a) \cdot m^{2}}{2 a+b}\right]=\frac{2}{2 a+b}+\frac{2}{a} \cdot \frac{2 b}{2 a+b} \cdot m^{2} \geqslant \frac{2}{2 a+b}$, hence the range of $f$ is $\left[\frac{2}{2 a+b},+\infty\right)$. Combining (1) and (2) we get: When $b<2 a$, the range of $f=x^{2}+y^{2}$ is
$$
\left[\frac{2}{2 a+b}, \frac{2}{2 a-b}\right]
$$

When $b \geqslant 2 a$, the range of $f=x^{2}+y^{2}$ is $\left[\frac{2}{2 a+b},+\infty\right)$.