高数中求偏导的题目，z^3-3xyz=a^3, 求d^2z/dxdy, 的二次混合求导

用隐函数微分法
令F[x,y,z] = z³-3xyz-a³
z'x = -F'x/F'z = yz/(z²-xy)
z'y = -F'y/F'z = xz/(z²-xy)
(z也是y的函数，刚才我当成常数扔了- -!)
z''xy = [z'x]'y = [(yz)'(z² - xy) - yz * (2z z'y - x)]/(z²-xy)²
= [(z + y z'y)(z²-xy) - 2yz² z'y + xyz]/(z²-xy)²
= (z³ - yz² z'y - xy² z'y)/(z²-xy)²
= [z³ - (yz²+xy²)xz/(z²-xy)]/(z²-xy)²
= z(z^4 - 2xyz³ - x²y²z)/(z²-xy)³

3z^2*dz/dx - 3yz - 3xy*dz/dx = 0,

(z^2-xy)dz/dx - yz = 0,dz/dx = yz/(z^2 - xy),

3z^2*dz/dy - 3xz - 3xy*dz/dy = 0,
(z^2 - xy)dz/dy - xz = 0, dz/dy = xz/(z^2 - xy),
(2z*dz/dx - y)dz/dy + (z^2 - xy)d^2z/(dxdy) - z - x*dz/dx = 0,

0 = [2z*yz/(z^2-xy) - y]xz/(z^2-xy) + (z^2-xy)d^2z/(dxdy) - z - xyz/(z^2-xy) = xyz[z^2 + xy]/(z^2-xy)^2 + (z^2-xy)d^2z/(dxdy) - z^3/(z^2-x