设F(x,y)= x2/3 + y2/3 -1
先对x求偏导数 再对 y求偏导数 在（x0,y0)点处的偏导数
Fx(x,y)=(2/3)x0^(-1/3)
Fy(x,y)=(2/3)y0^(-1/3)
切线就是 (2/3)x0^(-1/3)(x-x0)+(2/3)y0^(-1/3)(y-y0)=0
x0^(-1/3)(x-x0)+y0^(-1/3)(y-y0)=0
第一象限 x0 ,y0>0
求xy轴的正半轴交点 (0,x0^(2/3)y0^(1/3)+y0)
(y0^(2/3)x0^(1/3)+x0,0)
求距离 ={^2+^2}^(1/2)
x0,y0 还满足方程x2/3 + y2/3 =1 x0^(2/3)+y0^(2/3)=1
距离 ={^2+^2}^(1/2)
={^2+^2}^(1/2)
={^2+^2}^(1/2)
={x0^(2/3)+y0^(2/3)}^(1/2)
=1