[(x/y)e^(x/y)+1]dx/dy-(x/y)^2e^(x/y)=0
令x/y=u,则x=uy,dx/dy=ydu/dy+u
所以(ue^u+1)(ydu/dy+u)-u^2e^u=0
(ue^u+1)ydu/dy+u^2e^u+u-u^2e^u=0
(ue^u+1)ydu/dy=-u
(e^u+1/u)du=-dy/y
两边积分：e^u+ln|u|=-ln|y|+C
即e^(x/y)+ln|x|=C