Answer
$dx/dy=\frac{2x^4y-x^3+6xy^2}{4x^3y^2-3x^2y+2y^3}$
Work Step by Step
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:
$dx/dy=\frac{2x^4y-x^3+6xy^2}{4x^3y^2-3x^2y+2y^3}$
