Answer:
[tex]\dfrac{dy}{dx}=\dfrac{-4x^3y+2y^3+3x^2y^3}{x^4-6xy^2-3x^3y^2}[/tex]
Step-by-step explanation:
Normally, when I do this, I differentiate each term first with respect to x then with respect to y. In this solution, I differentiated the entire expression with respect to x, then with respect to y. That makes separating the dx and dy coefficients much easier, so the solution almost falls into your lap.
[tex]-x^4 y + 2 x y^3 + x^3 y^3=0 \qquad\text{given}\\\\(-4x^3y+2y^3+3x^2y^3)dx+(-x^4+6xy^2+3x^3y^2)dy=0\\\\(-4x^3y+2y^3+3x^2y^3)dx=(x^4-6xy^2-3x^3y^2)dy\\\\\boxed{\dfrac{dy}{dx}=\dfrac{-4x^3y+2y^3+3x^2y^3}{x^4-6xy^2-3x^3y^2}}[/tex]
