y=2x/(x-1)²
y'=[2(x-1)²-2x*2(x-1)]/(x-1)^4
=(2x²-4x+2-4x²+2x)/(x-1)^4
=(-2x²-2x+2)/(x-1)^4
=-2(x²+x-1)/(x-1)^4
y=2/x²+1
k=y'=2*(-2)x^(-3)+1
当x=1时,k=4+1=5
设切线方程为：y=5x+m
切线过（1,1）,则m=-4
所以曲线y=2/x²+1在点P（1,1）处的切线方程是：y=5x-4