Answer
$ \frac{dy}{dx} = \frac{x^2-2y}{2x+y^2}$
Work Step by Step
Applying the rules of derivatives:
$x^3-y^3 = 6xy$
$\frac{d}{dx} [x^3-y^3 ] = \frac{d}{dx} [6xy]$
$3x^2 - 3y^2 \frac{dy}{dx} = 6y + 6x * \frac{dy}{dx}$
$ \frac{dy}{dx} (6x+3y^2) = 3x^2-6y$
$ \frac{dy}{dx} = \frac{3x^2-6y}{6x+3y^2} = \frac{x^2-2y}{2x+y^2}$
