# How do you determine dy/dx given x^2+xy-y^3=xy^2?

Nov 13, 2016

$\left\{\begin{matrix}x + y = 0 \to \frac{\mathrm{dy}}{\mathrm{dx}} = - 1 \\ x - {y}^{2} = 0 \to \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{x}}\end{matrix}\right.$

#### Explanation:

${x}^{2} + x y - {y}^{3} - x {y}^{2} = \left(x + y\right) \left(x - {y}^{2}\right) = 0$ This is equivalent to

$\left\{\begin{matrix}x + y = 0 \\ x - {y}^{2} = 0\end{matrix}\right.$

Solving we have

$\left\{\begin{matrix}x + y = 0 \to \frac{\mathrm{dy}}{\mathrm{dx}} = - 1 \\ x - {y}^{2} = 0 \to \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{x}}\end{matrix}\right.$