## Calculus: Early Transcendentals 8th Edition

$dx/dy=\frac{2x^4y-x^3+6xy^2}{4x^3y^2-3x^2y+2y^3}$
Take the derivative as is on either side of the equation: $2x^4y+4x^3y^2(dx/dy)-x^3-3x^2y(dx/dy)+6xy^2+2y^3(dx/dy)=0$ Isolate all terms with dx/dy onto one side of the equal sign and distribute the dx/dy out of each term: $dx/dy(4x^3y^2-3x^2y+2y^3)=2x^4y-x^3+6xy^2$ Isolate dx/dy by dividing both sides by the terms: $dx/dy=\frac{2x^4y-x^3+6xy^2}{4x^3y^2-3x^2y+2y^3}$