Answer
$\dfrac{dy}{dx}=\dfrac{3y^3x^2-1}{1-3x^3y^2}.$
Work Step by Step
$\dfrac{d}{dx}(x^3y^3)-\dfrac{d}{dx}(y)=\dfrac{d}{dx}(x)\rightarrow$
$(3x^2y^3+\dfrac{dy}{dx}(3y^2x^3))-\dfrac{dy}{dx}=1\rightarrow$
$3y^3x^2-1=\dfrac{dy}{dx}(1-3x^3y^2)\rightarrow$
$\dfrac{dy}{dx}=\dfrac{3y^3x^2-1}{1-3x^3y^2}.$
