设x/z=ln(z/y),求d^2z/dx^2和d^2z/dy^2 用多元函数方法求隐函数的二阶导数

F(x,y,z)=x/z-ln(z/y)=0
Fx+Fz*dz/dx=0 => dz/dx=-Fx/Fz
Fy+Fz*dz/dy=0 => dz/dy=-Fy/Fz
Fx=1/z,
Fy=-y/z*(-z/y^2)=1/y
Fz=-x/z^2-y/z*(1/y)=-x/z^2-1/z
dz/dx=-Fx/Fz=(-1/z)/(-x/z^2-1/z)=z/(x+z)
d^2z/dx^2=d(dz/dx)/dx=d(-Fx/Fz)/dx+d(-Fx/Fz)/dz*dz/dx
=-z/(x+z)^2+[(x+z)-z]/(x+z)^2*z/(x+z)
=z/(x+z)^2*[x/(x+z)-1]
=-z^2/(x+z)^3
dz/dy=-Fy/Fz=(-1/y)/(-x/z^2-1/z)=z/y*z/(x+z)=z^2/[y(x+z)]
d^2z/dy^2=d(dz/dy)/dy=d(-Fy/Fz)/dy+d(-Fy/Fz)/dz*dz/dy
=z^2/(x+z)*(-1/y^2)+1/y*[2z(x+z)-z^2]/(x+z)^2*z^2/[y(x+z)]
=-z^2/[y^2(x+z)]+z^2/[y^2(x+z)]*(z^2+2xz)/(x+z)^2
=z^2/[y^2(x+z)]*[(z^2+2xz)/(x+z)^2-1]
=-x^2z^2/[y^2(x+z)^3]
1/z+(-x/z^2)*dz/dx=1/zdz/dx