[Sirius]

จาก $\frac{a}{b} +\frac{b}{a}=6$ ดังนั้น $\frac{a^2+b^2}{ab}=6$
$\frac{ab}{a^2+b^2}=\frac{1}{6}$
$(\frac{ab}{a^2+b^2})^2=\frac{1}{36}$
$\frac{a^2b^2}{a^4+2a^2b^2+b^4}=\frac{1}{36}$
$\therefore (\frac{a^2-b^2}{a^2+b^2})^2=\frac{a^4-2a^2b^2+b^4}{a^4+2a^2b^2+b^4}$
$=\frac{a^4+2a^2b^2+b^4}{a^4+2a^2b^2+b^4}-\frac{4a^2b^2}{a^4+2a^2b^2+b^4}$
$=1-\frac{4}{36}=\frac{8}{9}$