x-y+z=e^z
对x求导：1+z'x=z'x*e^z,得 z'x=1/(e^z-1)
对y求导：-1+z'y=z'y*e^z,得：z'y=-1/(e^z-1)
因此dz=z'xdx+z'ydy=(dx-dy)/(e^z-1)