$$x^2y^3 + xy^2 + y = 3$$
I am supposed to find the equation of the tangent to this curve. I have gotten my implicit differential answer but I dont think that I am correct. I know the method and concept of doing this but I just wish to confirm my answers here. I got:
$y^3 + 2x$ as the differential (derivative)....is that correct?
$x^2y^3 + xy^2 + y = 3\\ \frac {d}{dx}(x^2y^3 + xy^2 + y) = \frac {d}{dx}3\\ (2xy^3 + y^2) + (3x^2y^2 + 2xy + 1)\frac {dy}{dx} = 0\\ \frac {dy}{dx} = -\frac {2xy^3 + y^2}{3x^2y^2 + 2xy + 1}$