如图
x=yln(xy)，等式两端对x求导，1=dy/dx+y[1/ln(xy)][y+x(dy/dx)]=dy/dx+y/ln(xy)+xdy/dx整理得(dy/dx)(1+x)=1－y/ln(xy)，即y'=dy/dx={[ln(xy)-y]/[(1+x)ln(xy)]
