Answer
$\dfrac{dy}{dx}=\dfrac{3x^2-6xy+2y^2}{3x^2-4yx}.$
Work Step by Step
$\dfrac{d}{dx}(x^3)-\dfrac{d}{dx}(3x^2y)+\dfrac{d}{dx}(2y^2x)=\dfrac{d}{dx}(12)\rightarrow$
$3x^2-(6xy+\dfrac{dy}{dx}(3x^2))+(\dfrac{dy}{dx}(4yx)+2y^2)=0\rightarrow$
$3x^2-6xy+2y^2=\dfrac{dy}{dx}(3x^2-4yx)\rightarrow$
$\dfrac{dy}{dx}=\dfrac{3x^2-6xy+2y^2}{3x^2-4yx}.$
