两边求导， x'y+xy'-1/y*y'=0即y+xy'-y'/y=0求得 y'=y/(1/y-x)=y^2/(1-xy)
xy-lny=3y+xy'-y'/y=0y=(1/y-x)y'y'=y/[(1-xy)/y]=y^2/(1-xy)注：^2——表示平方。
