Answer
$\frac{dy}{dx}= \frac{y^2-x^2}{y^2-2xy}$
Work Step by Step
$x^3+y^3= 3xy^2$
$ \frac{d}{dx} ( x^3+y^3)= \frac{d}{dx}(3xy^2))$
$ \frac{d}{dx}(x^3)+\frac{d}{dx}y^3=3\frac{d}{dx} (3xy^2) $
$3x^2 + \frac{d}{dy}y^3\frac{dy}{dx}=3[x\frac{d}{dx} y^2+y^2\frac{dx}{dx}] $
$3x^2 + 3y^2\frac{dy}{dx}=3[x\frac{d}{dy} y^2 \frac{dy}{dx}+y^2] $
$3x^2 + 3y^2\frac{dy}{dx}=3[x(2y) \frac{dy}{dx}+y^2] $
$3x^2+3y^2\frac{dy}{dx} =6xy\frac{dy}{dx}+3y^2$
$(3y^2-6xy) \frac{dy}{dx}=3y^2-3x^2$
$3(y^2-2xy) \frac{dy}{dx}=3(y^2-x^2)$
$\frac{dy}{dx}= \frac{y^2-x^2}{y^2-2xy}$
