一道数学题，帮忙解答一下！谢谢！

已知x+1/x=3，求x^2/(x^4+x^2+1)的值！我不太会做，请大家帮帮忙，要有过程谢谢！！

还有一道：先化简后求值！
[4/(x^2-y^2)+(x+y)/(x*y^2-x^2*y)]/[(x^2+xy-2*y^2)/x^2*y+2x*y^2)]
其中x=2,y=-3，请写上过程，谢谢！！！！！！！！！！
非常感谢！！！！！！！！！

x+1/x=3

把x^2/[x^4+x^2+1]倒过来是[x^4+x^2+1]/x^2=x^2+1/x^2+1=[x+1/x]^2-2+1=3*3-1=8

所以：x^2/[x^4+x^2+1]=1/8

[4/(x^2-y^2)+(x+y)/(x*y^2-x^2*y)]/[(x^2+xy-2*y^2)/x^2*y+2x*y^2)]
=[4/(x+y)(x-y)+(x+y)/xy(y-x)]/[(x+2y)(x-y)/xy(x+2y)]
=[(4xy-x^2-2xy-y^2)/xy(x+y)(x-y)]/[(x-y)/xy]
=[-(x-y)^2/xy(x+y)(x-y)]/[(x-y)/xy]
=-[x-y]/xy(x+y)*xy/[x-y]
=-1/[x+y]
=-1/[2-3]
=1

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