对x求导有：
e^(x+y)*(1+y')+cos(xy)(y+xy')=1
[e^(x+y)+cos(xy)x]y'=1-e^(x+y)-y*cos(xy)
y'=[1-e^(x+y)-y*cos(xy)] / [e^(x+y)+cos(xy)x]
e^(x+y)+sin(xy)=x
(1+y')e^(x+y)+[cos(xy)](y+xy')=1
y'=-[e^(x+y)+ycos(xy)-1]/[e^(x+y)+xcos(xy)]
两边关于x求导
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