a^2-3a+1=0
a^2=3a-1
显然a≠0,两边同除以a,有：
a+1/a=3
两边再平方,有：
a^2+1/a^2+2=9
a^2+1/a^2=7
设y=a^2/(a^6+1)
则 1/y=a^4+1/a^2
=(3a-1)^2+7-a^2
=8a^2-6a+8
=8(3a-1)-6a+8
=18a
所以 y=1/(18a)
1/a=3-a
y=(3-a)/18
a^2-3a+1=0
解得 a=(3±√5)/2
y=(3-(3±√5)/2)/18
=(3±√5)/18