Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 23

Find $\displaystyle \frac{dx}{dy}$ of $x^4y^2 - x^3y + 2xy^3 = 0$ by Implicity Differentiation.
$\displaystyle \frac{d}{dy} (x^4y^2) - \frac{d}{dy} (x^3y) + \frac{d}{dy} (2xy^3) =0$


$
\begin{equation}
\begin{aligned}
\left[ (x^4) \frac{d}{dy} (y^2) + (y^2) \frac{d}{dy} (x^4) \right] - \left[ (x^3) \frac{d}{dy} (y) + (y) \frac{d}{dy} (x^3) \right] +2 \left[ (x) \frac{d}{dy} (y^3) + (y^3) \frac{d}{dy} \right] &= 0\\
\\
\left[ (x^4) (2y) + (y^2)(4x^3) \frac{dx}{dy}\right] - \left[ (x^3)(1)+(y)(3x^2) \frac{dx}{dy}\right] +2 \left[ (x)(3y^2)+(y^3) \frac{dx}{dy}\right] &= 0\\
\\
2x^4y + 4x^3y^2 \frac{dx}{dy} - x^3 - 3x^2y \frac{dx}{dy} + 6xy^2 + 2y^3 \frac{dx}{dy} &= 0\\
\\
4x^3y^2 \frac{dx}{dy} - 3x^2y \frac{dx}{dy} + 2y^3 \frac{dx}{dy} &= x^3 - 2x^4y - 6xy^2\\
\\
\frac{dx}{dy} ( 4x^3 y^2 - 3x^2y + 2y^3) &= x^3 - 2x^4y -6xy^2\\
\\
\frac{\frac{dx}{dy} \cancel{( 4x^3 y^2 - 3x^2y + 2y^3)} }{\cancel{( 4x^3 y^2 - 3x^2y + 2y^3)}} &= \frac{x^3-2x^4y-6xy^2}{4x^3y^2-3x^2y+2y^3}\\
\\
\frac{dx}{dy} & = \frac{x^3-2x^4y-6xy^2}{4x^3y^2-3x^2y+2y^3}
\end{aligned}
\end{equation}
$