思路：x+y=e^xy ,两边取微分d(x+y)=d(e^xy)dx+dy=e^xyd(xy)dx+dy=e^xy(xdy+ydx)dx+dy=xe^xydy+ye^xydx(xe^xy-1)dy=(1-ye^xy)dxdy/dx=[(1-ye^xy)/(xe^xy-1)]dx代入x+y=e^xy,得dy/dx=[1-y(x+y)]/[x(x+y)-1]=(1-y²...