xy3/(1+y2).
Let u=xy3 and v=(1+y2)-1.
u'=3xy2y'+y3; v'=-(1+y2)-2(2yy'), where y'=dy/dx.
d(uv)/dx=vu'+uv'=(3xy2y'+y3)/(1+y2)-xy3(1+y2)-2(2yy')=(3xy2y'+y3)/(1+y2)-2xy4y'/(1+y2)2.
If z=uv, then z'=(3xy2y'+y3)/(1+y2)-2xy4y'/(1+y2)2 is the differential of the given expression.
Since we are not given the relationship between y and x, this differential has to include the unknown y'.