Answer
$\Large\frac{dx}{dy}=\frac{-2x^4y+x^3-6xy^2}{4x^3y^2-3x^2y+2y^3}$
Work Step by Step
$x^4y^2-x^3y+2xy^3=0$ ____(1)
Differentiating (1) with respect to $y$ treating $x$ as dependent variable and $y$ as independent variable
$\left[4x^3\frac{dx}{dy}y^2+2x^4y\right]-\left[3x^2\frac{dx}{dy}y+x^3\right]+\left[2\frac{dx}{dy}y^3+6xy^2\right]=0$
$\frac{dx}{dy}[4x^3y^2-3x^2 y+2y^3]=-2x^4y+x^3-6xy^2$
$\Large\frac{dx}{dy}=\frac{-2x^4y+x^3-6xy^2}{4x^3y^2-3x^2y+2y^3}$
Hence $\Large\frac{dx}{dy}=\frac{-2x^4y+x^3-6xy^2}{4x^3y^2-3x^2y+2y^3}$.
