求方程xyz + x2 + y2 + z2 = 2 确定的函数z = z( x,y)在点(1,0,-1)处的全微分dz,

为方便,记p=√(x^2+y^2+z^2)
对x求导：yz+xyz'x+(x+zz'x)/p=0, 得：z'x=-(yz+x/p)/(xy+z/p)
同样,对y求导,得：z'y=-(xz+y/p)/(xy+z/p)
所以在(1,0,-1)处,
有p=√2
z'x=-(1/p)/(-1/p)=1
z'y=-(-1)/(-1/p)=-1/p=-√2/2
所以dz=z'xdx+z'ydy=dx-√2/2*dy