Answer
$y'=\dfrac{y^{2}-3x^{2}}{3y^{2}-2xy}$
Work Step by Step
$x^{3}-xy^{2}+y^{3}=1$
Differentiate each term:
$(x^{3})'-(xy^{2})'+(y^{3})'=(1)'$
Use the product rule to find $(xy^{2})'$:
$3x^{2}-[(x)(y^{2})'+(y^{2})(x)']+(3y^{2})(y')=0$
$3x^{2}-[(x)(2y)(y')+(y^{2})(1)]+(3y^{2})(y')=0$
$3x^{2}-2xyy'-y^{2}+3y^{2}y'=0$
Solve for $y'$:
$-2xyy'+3y^{2}y'=-3x^{2}+y^{2}$
$y'(-2xy+3y^{2})=-3x^{2}+y^{2}$
$y'=\dfrac{-3x^{2}+y^{2}}{-2xy+3y^{2}}=\dfrac{y^{2}-3x^{2}}{3y^{2}-2xy}$
