Given: #2 = e^(xy^2-x^3y) - y^2x^3 + y#
Derivative rules:
#(k)' = 0#
#(e^u)' = u'e^u#
Product rule: #(uv)' = uv' + vu'#
Use the product rule twice for #e#:
Let #u = xy^2 - x^3y#
#u' = 2xyy' + y^2 -x^3y' -3x^2y#
#(e^u)' = (2xyy' + y^2 -x^3y' -3x^2y)e^(xy^2-x^3y)#
#(-y^2x^3)' = -3x^2y^2 -2x^3yy'#
Put it all together
#0 = (2xyy' + y^2 -x^3y' -3x^2y)e^(xy^2-x^3y) -3x^2y^2 - 2x^3yy' + y'#
Distribute the #e#:
#0 = 2xyy'e^(xy^2-x^3y) + y^2e^(xy^2-x^3y)-x^3y'e^(xy^2-x^3y) -3x^2y e^(xy^2-x^3y) -3x^2y^2 - 2x^3yy' + y'#
Move all the #y'# terms to the left side:
#-2xyy'e^(xy^2-x^3y) + x^3y'e^(xy^2-x^3y) +2x^3yy' - y' = y^2e^(xy^2-x^3y)-3x^2y e^(xy^2-x^3y) -3x^2y^2#
Factor the #y'#:
#y'(-2xye^(xy^2-x^3y) + x^3e^(xy^2-x^3y) +2x^3y - 1) = y^2e^(xy^2-x^3y)-3x^2y e^(xy^2-x^3y) -3x^2y^2#
Divide:
#y' = (y^2e^(xy^2-x^3y)-3x^2y e^(xy^2-x^3y) -3x^2y^2)/((-2xye^(xy^2-x^3y) + x^3e^(xy^2-x^3y) +2x^3y - 1))#
Factor out a negative from the numerator and rearrange:
#y' = -(3x^2y^2 + 3x^2y e^(xy^2-x^3y) -y^2e^(xy^2-x^3y))/(2x^3y - 1 -2xye^(xy^2-x^3y) + x^3e^(xy^2-x^3y))#
Make #e^(xy^2-x^3y) = e^(xy^2)e^(-x^3y)#
#y' = -(3x^2y^2 + 3x^2y e^(xy^2)e^(-x^3y) -y^2e^(xy^2)e^(-x^3y))/(2x^3y - 1 -2xye^(xy^2)e^(-x^3y) + x^3e^(xy^2)e^(-x^3y))#
Factor #ye^(-x^3y)# from both numerator and #e^(-x^3y)# from the denominator.
Realize that #e^(-x^3y)e^(x^3y) = e^0 = 1#
#y' = -(ye^(-x^3y)(3x^2ye^(x^3y) + 3xe^(xy^2) - ye^(xy^2)))/(e^(-x^3y)(2x^3ye^(x^3y) - e^(x^3y) - 2xye^(xy^2)+x^3e^(xy^2)))#
Cancel common factors:
#y' = -(y(3x^2ye^(x^3y)+3xe^(xy^2)-ye^(xy^2)))/(2x^3ye^(x^3y)-e^(x^3y)-2xye^(xy^2)+x^3e^(xy^2)#
