13. Let \( f(x, y) = \frac{a x^{2} + x y + y^{2}}{x^{2} + y^{2}} \), satisfying
\[
\max _{x^{2}+y^{2}+0} f(x, y) - \min _{x^{2}+y^{2}+0} f(x, y) = 2 \text{. }
\]

Find \( a \).
$$
\begin{array}{l}
=\max _{0 \leqslant \theta<2 \pi}\left(\frac{a-1}{2} \cos 2 \theta+\frac{1}{2} \sin 2 \theta+\frac{a+1}{2}\right) \\
=\max _{0 \leqslant \theta<2 \pi}\left(\sqrt{\left(\frac{a-1}{2}\right)^{2}+\frac{1}{4}} \sin (2 \theta+a)+\frac{a+1}{2}\right) \\
=\sqrt{\left(\frac{a-1}{2}\right)^{2}+\frac{1}{4}}+\frac{a+1}{2} .
\end{array}
$$

Similarly, $\max _{x^{2}+y^{2} y_{0}} f(x, y)=-\sqrt{\left(\frac{a-1}{2}\right)^{2}+\frac{1}{4}}+\frac{a+1}{2}$. Therefore, $\max _{x^{2}+y^{2}+0} f(x, y)-\min _{x^{2}+y^{2} \neq 0} f(x, y)$
$$
=2 \sqrt{\left(\frac{a-1}{2}\right)^{2}+\frac{1}{4}} \text {. }
$$

Thus, $(a-1)^{2}=3$, which means $a=1 \pm \sqrt{3}$. Three, 13. Since $f(k x, k y)=f(x, y)$, we have
$$
\begin{array}{l}
\max _{x^{2}+y^{2}} f(x, y)=\max _{x^{2}+y^{2}=1} f(x, y) \\
=\max _{0<\theta<2 \pi}\left(a \cos ^{2} \theta+\cos \theta \cdot \sin \theta+\sin ^{2} \theta\right) \\
=\max _{0<\theta<2 \pi}\left((a-1) \cos ^{2} \theta+\cos \theta \cdot \sin \theta+1\right)
\end{array}
$$