Answer
$\dfrac{dy}{dx}=\dfrac{4}{5}$
Work Step by Step
$\dfrac{d}{dx}(x^3)+\dfrac{d}{dx}(y^3)-\dfrac{d}{dx}(6xy)=0\rightarrow$
$3x^2+\dfrac{dy}{dx}(3y^2)-(6y+\dfrac{dy}{dx}(6x))=0\rightarrow$
$3x^2-6y=\dfrac{dy}{dx}(6x-3y^2)\rightarrow$
$\dfrac{dy}{dx}=\dfrac{x^2-2y}{2x-y^2}.$
At $(\dfrac{4}{3}, \dfrac{8}{3})\rightarrow\dfrac{dy}{dx}=\dfrac{(\frac{4}{3})^2-2(\frac{8}{3})}{2(\frac{4}{3})-(\frac{8}{3})^2}=\dfrac{4}{5}.$
