Correct option is (b) \(\frac{(3x^2-3y)}{3x-3y^2-2y}\)
The explanation is: Differentiation of x^3 is 3x^2
differentiation of y^3 is 3y^2 \(\frac{dy}{dx}\)
differentiation of -3xy is [-3y -3x \(\frac{dy}{dx}\)]
differentiation of y^2 is 2y \(\frac{dy}{dx}\)
Hence,
\(\frac{d(x^3+y^3-3xy+y^2)}{dx}=0\)
\(3x^2+3y^2 \frac{dy}{dx}-3y-3x \frac{dy}{dx}+2y \frac{dy}{dx} = 0\)
\(\frac{dy}{dx}=\frac{(3x^2-3y)}{3x-3y^2-2y}\)