Answer
$y'=-\dfrac{4x^{3}+2xy^{2}}{2x^{2}y+3y^{2}}$
Work Step by Step
$x^{4}+x^{2}y^{2}+y^{3}=5$
Differentiate each term:
$(x^{4})'+(x^{2}y^{2})'+(y^{3})'=(5)'$
Use the product rule to find $(x^{2}y^{2})'$:
$4x^{3}+(x^{2})(y^{2})'+(y^{2})(x^{2})'+(3y^{2})(y')=0$
$4x^{3}+(x^{2})(2y)(y')+(y^{2})(2x)+(3y^{2})(y')=0$
Solve for $y'$:
$2x^{2}yy'+3y^{2}y'=-4x^{3}-2xy^{2}$
$y'(2x^{2}y+3y^{2})=-4x^{3}-2xy^{2}$
$y'=\dfrac{-4x^{3}-2xy^{2}}{2x^{2}y+3y^{2}}=-\dfrac{4x^{3}+2xy^{2}}{2x^{2}y+3y^{2}}$
