两边求导(2/3)x^(-1/3)+(2/3)y^(-1/3)y'=0得y'=-(x/y)^(-1/3)不妨设切点为(x0,y0),则切线斜率k=-(x0/y0)^(-1/3)切线方程为y-y0=-(x0/y0)^(-1/3)(x-x0)计算得切线与坐标轴交于点(0,y0^(1/3)a^(2/3)),点(x0^(1/3)a^(2/...