# How do you find (dy)/(dx) given 4x^2+3xy^2-6x^2y=y^3?

Sep 1, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 8 x - 3 {y}^{2} + 12 x y}{6 x y - 6 {x}^{2} - 3 {y}^{2}}$

#### Explanation:

Given -

$4 {x}^{2} + 3 x {y}^{2} - 6 {x}^{2} y = {y}^{3}$

$8 x + \left[\left(3 x \frac{\mathrm{dy}}{\mathrm{dx}} 2 y\right) + \left({y}^{2.} 3\right)\right] - \left[\left(6 {x}^{2} . \frac{\mathrm{dy}}{\mathrm{dx}}\right) + \left(y . 12 x\right)\right] = \frac{\mathrm{dy}}{\mathrm{dx}} 3 {y}^{2}$

$8 x + \left[6 x y \frac{\mathrm{dy}}{\mathrm{dx}} + 3 {y}^{2}\right] - \left[6 {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + 12 x y\right] = \frac{\mathrm{dy}}{\mathrm{dx}} 3 {y}^{2}$

$8 x + 6 x y \frac{\mathrm{dy}}{\mathrm{dx}} + 3 {y}^{2} - 6 {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} - 12 x y = \frac{\mathrm{dy}}{\mathrm{dx}} 3 {y}^{2}$

$8 x + 6 x y \frac{\mathrm{dy}}{\mathrm{dx}} + 3 {y}^{2} - 6 {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} - 12 x y - \frac{\mathrm{dy}}{\mathrm{dx}} 3 {y}^{2} = 0$

$6 x y \frac{\mathrm{dy}}{\mathrm{dx}} - 6 {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{\mathrm{dy}}{\mathrm{dx}} 3 {y}^{2} = - 8 x - 3 {y}^{2} + 12 x y$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(6 x y - 6 {x}^{2} - 3 {y}^{2}\right) = - 8 x - 3 {y}^{2} + 12 x y$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 8 x - 3 {y}^{2} + 12 x y}{6 x y - 6 {x}^{2} - 3 {y}^{2}}$