为自己加油

设e^z-xyz=0，求∂z/∂x²，∂²z/∂y²，∂²z/∂x∂y．

设F(x，y, z)=e^z-xyz，则： F_x=-yz，F_y=-xz，F_z=e^z-xy ∴∂z/∂x=-(F_x/F_z)=yz/(e^z-xy) ∂z/∂y=-(F_y/F_z)=xz/(e^z-xy) ∂²z/∂x²=∂/∂x[yz(e^z-xy)]=[z′(e^zxy)-yz(e^zz′-xy)]/ (e^z-xy) =(2y²ze^z-2y³xz-y²z²e^z)/(e^z-xy)³= (z³-2z²+2z)/ [x²(1-z)³] ∂²z/∂y²=∂/∂y[xz(e^zxy)]= (2x²ze^z-2x³yz-x²z²e^z)/(e^z-xy)³= (z³-2z²+2z)/ [y²(1-z)³] ∂²z/∂x∂y=∂/∂y[y^z/(e^z-xy)]=(ze^2z-xyz²e^z-x ²y²z)/(e^z-xy)³=z/[xy(1-z)³].