化为：e^(ylnx)-e^y=sin(xy)
两边对x求导：e^(ylnx)(y'lnx+y/x)-y'e^y=cos(xy)(y+xy')
y'[lnxe^(ylnx)-e^y-xcos(xy)]=[ycos(xy)-e^(ylnx)y/x]
y'=[ycos(xy)-x^y* y/x]/[lnx(x^y)-e^y-xcos(xy)]
因此dy=[ycos(xy)-x^y(y/x)]/[lnx(x^y)-e^y-xcos(xy)]dx