高数中一道计算二元偏导数的小题，我的计算与标答有所出入，在线等
设Z^3-3XYZ=a^3，求(dZ)^2/dXdY
备注：(dZ)^2/dXdY表示Z对X，Y二阶求导
答案解题过程：
设F(X,Y,Z)=Z^3-3XYZ-a^3,则
dZ/dX=-(-3YZ)/(3Z^2-3XY)=YZ/(Z^2-XY)
dZ/dY=-(-3XZ)/(3Z^2-3XY)=XZ/(Z^2-XY)
(dZ)^2/dXdY=d[YZ/(Z^2-XY)]/dY
=[(Z+Y*(dZ/dY))*(Z^2-XY)-YZ(2Z*(dZ/dY)-X)]/(Z^2-XY)^2
我的计算过程和标准答案的区别就在这里了，上面一步是标准答案，但我认为这一步应该是
=[(Z+Y*(dZ/dY))*(Z^2-XY)-YZ(2Z*(dZ/dY)-X-Y*(dX/dY))]/(Z^2-XY)^2
所以我接下来的结果就是
=[Z^3-XYZ+XYZ^3/(Z^2-XY)-(X^2)*(Y^2)*Z/(Z^2-XY)-YZ*(2X*(Z^2)/(Z^2-XY))]/(Z^2-XY)^2
我错了吗？？？？
[解决办法]
Z^3-3XYZ=a^3
0=F(x,y,z)=z^3-3xyz
df/dx = 3z^2*dz/dx - 3yz-3xydz/dx=0
df/dy = 3z^2*dz/dy - 3xz-3xydz/dy=0
dz/dx = 3yz/(3z^2-3xy)=yz/(z^2-xy)
dz/dy = xz/(z^2-xy)
dz^2/dxdy = d(dz/dx)/dy = d(yz/(z^2-xy))/dy
其中：d(z^2-xy)/dy = 2zdz/dy - x = 2xz^2/(z^2-xy)-x
d(yz)/dy = z+ydz/dy = z + xyz/(z^2-xy)
所以：dz^2/dxdy = d(dz/dx)/dy = d(yz/(z^2-xy))/dy
=[(z^2-xy)d(yz)/dy - yzd(z^2-xy)/dy]/(z^2-xy)^2
=[(x^2-xy)(z+xyz/(z^2-xy))-yz(2xz^2/(z^2-xy) -x )]/(z^2-xy)^2
你的式子没错，但是要记得x，y在这里是无关的两个自变量，dx/dy和dy/dx都是0
