xy∧2+lny+sin(3x+2y)=1
等式两边对x求导，
y^2+2xy*(dy/dx)+(dy/dx)/y+cos(3x+2y)(3+2(dy/dx))=0
(2xy+1/y+2cos(3x+2y))(dy/dx)=-y^2-3cos(3x+2y)
dy/dx=(-y^2-3cos(3x+2y))/((2xy+1/y+2cos(3x+2y))
y^2+2xyy'+y'/y+cos(3x+2y)(3+2y')=0
y'=-[y^3+3ycos(3x+2y)]/[2xy^2+2ycos(3x+2y)+1]
dy/dx=-[y^3+3ycos(3x+2y)]/[2xy^2+2ycos(3x+2y)+1]
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