x²+y²-xy=1 等式两边对x求一阶导 得2x+2yy‘-y-xy’=0 得y'=（2x-y）/（x-2y）2x+2yy‘-y-xy’=0 等式两边对x求导 得2+2y'²+2yy‘’-y‘-y’-xy‘’=0 得y‘’=（2+2y'²-2y'）/（x-2y）将y'=（2...