If x^3+y^3=xy, then dy/dx is equal to :

The correct answer is Option (2) → $\frac{y-3x^2}{3y^2-x}$

$x^3+y^3=xy$

Differentiate w.r.t 'x'

$3x^2+3y^2\frac{dy}{dx}=y+x\frac{dy}{dx}$

$⇒\frac{dy}{dx}(3y^2-x)=y-3x^2$

$⇒\frac{dy}{dx}=\frac{y-3x^2}{3y^2-x}$