解：du=fxdx+fydy+fzdz 根据微分不变性 ……A右边，(xe^x+e^x)dx-(ye^y+e^y)dy=(ze^z+e^z)dz即dz=((xe^x+e^x)dx-(ye^y+e^y)dy)/(ze^z+e^z)带入A整理du/=(fx+fz(xe^x+e^x)/(ze^z+e^z))dx+(fy-(ye^y+e^y)/(ze^z+e^z))dy
