设x+y+z=0,x平方+y平方+z平方=1求y对x全导以及z对x全导

两个方程，三个变量，因此自变量一个，函数有两个，由题意，x是自变量，因此y、z均是函数。
两方程两边分别对x求导得：
1+y'+z'=0
2x+2yy'+2zz'=0
即：
1+y'+z'=0
(1)
x+yy'+zz'=0
(2)
(1)×y-(2)得：
y-x+(y-z)z'=0
解得：z'=(x-y)/(y-z)

x+y+z=0
1+dy/dx + dz/dx =0 (1)

x^2+y^2+z^2 =1
2x+ 2y.dy/dx + 2z.dz/dx = 0

2x+ 2(-x-z).(-1-dz/dx) + 2z.dz/dx = 0
x+ (x+z).(1 +dz/dx) + z.dz/dx = 0
x + x+ x.dz/dx +z + z.dz/dx +z.dz/dx = 0
(x+2z).dz/dx =-(2x+z)
dz/dx =-(2x+z)/(x+2z)
--------------
x^2+y^2+z^2 =1
2x+ 2y.dy/dx + 2z.dz/dx = 0

2x+ 2y.dy/dx + 2(-x-y).(-1-dy/dx) = 0
x+ y.dy/dx + (x+y).(1+dy/dx) = 0
x+ y.dy/dx + x+x.dy/dx +y +y.dy/dx =0
(x+2y).dy/dx =-(2x+y)

dy/dx =-(2x+y)/(x+2y)

这不就出来了