Answer
$y'=\dfrac{(y-3x^2)}{(3y^2-x)}$
Work Step by Step
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, $y'=\dfrac{(y-3x^2)}{(3y^2-x)}$
