Answer
$y' = \frac{1-2xy - 6x^2y}{2x^3+x^2-1}$
Work Step by Step
$\frac{d}{dx}(x^2y) + \frac{d}{dx}(2x^3y) = \frac{d}{dx}(x) + \frac{d}{dx}(y)$
$(2xy + x^2y') + (2*3x^2y + 2x^3y') = 1 + y'$
Now we just need to simplify and solve for $y'$.
$(2xy + 6x^2y) + (x^2y' + 2x^3y') = 1 + y'$
$x^2y' + 2x^3y' -y'= 1 - 2xy - 6x^2y$
$y'(x^2 + 2x^3 -1) = 1 - 2xy - 6x^2y$
$y' = \frac{1-2xy - 6x^2y}{2x^3+x^2-1}$
