Find y' by implicit differentiation: xy + 2x + 3x^2 = 4

(xy) ' + (2x) ' + (3x^2) ' = (4) '

y + xy' + 2 + 6x = 0

xy' = - y - 2 - 6x

y' = [-y - 2 - 6x] / x

Now solve y from the original equation and substitue

xy + 2x + 3x^2 = 4 = > y = [-2x - 3x^2 + 4] / x

y' = [ (-2x - 3x^2 + 4) / x - 2 - 6x ] / x

y' = [-2x - 3x^2 + 4 - 2x - 6x^2 ] x^2 = [ - 4x - 9x^2 + 4] / x^2 =

= [-9x^2 - 4x + 4] / x^2