sin(xy)-ln[(x+1)/y]=1
两端对x求导得
cos(xy)*(xy'+y)-1/(x+1)+y'/y=0
再求导得
-sin(xy)*(xy'+y)+cos(xy)*(y'+xy''+y')+1/(x+1)^2+(y''y-y'^2)/y^2=0
自己整理吧.