Chapter 3: Derivatives - Section 3.7 - Implicit Differentiation - Exercises 3.7 - Page 155: 4

$y'=\dfrac{(y-3x^2)}{(3y^2-x)}$

Apply rule of differentiation: $y'=a'(x)+b'(x)$ Here, $a'(x)=m'(x)n(x)+m(x)n'(x)$ Given: $x^3-xy+y^3=\frac{1}{d/dx}$ $3x^2-(x'y+xy')+3y^2y'=0$ This implies, $3x^2-y-xy'+3y^2y'=0$ $y'(3y^2-x)=y-3x^2$ Hence, $y'=\dfrac{(y-3x^2)}{(3y^2-x)}$