求隐函数y=1-xe^y的导数解一：dy/dx=-e^y-x(e^y)(dy/dx),(1+xe^y)(dy/dx)=-e^y,故dy/dx=-(e^y)/(1+xe^y)；解二：作函数F(x,y)=y-1+xe^y≡0；则dy/dx=-(∂F/∂x)/(∂F/∂y)=-(e^y)/(1+xe^y)
