∵x³+3xy+y³=5 ==>3x²+3y+3xy'+3y²y'=0 (求导数)
==>y'=-(x²+y)/(x+y²)
∴切线在点(1,1)处的斜率是y'=-(1+1)/(1+1)=-1
故所求切线方程是y=-1*(x-1)+1,即y=2-x
所求法线方程是y=1*(x-1)+1,即y=x