Implicit differentiation of a mixed polynomial equation

Differentiate both sides with respect to $x$. For $y^4$, the chain rule gives $4y^3 y'$. For $x^2y^2$, apply the product rule: $\frac{d}{dx}[x^2 \cdot y^2] = 2x\cdot y^2 + x^2\cdot 2y y'$. The right side gives $3 + 2y'$. So the equation becomes $4y^3y' + 2xy^2 + 2x^2yy' = 3 + 2y'.$

Move all terms containing $y'$ to one side and everything else to the other: $4y^3y' + 2x^2yy' - 2y' = 3 - 2xy^2.$ Factor out $y'$: $y'(4y^3 + 2x^2y - 2) = 3 - 2xy^2.$ Dividing gives the final answer $y' = \dfrac{3 - 2xy^2}{4y^3 + 2x^2y - 2}.$

The product rule was used on the term x squared y squared, since it is a product of two functions of x. Differentiating it gave two x times y squared plus x squared times two y times dy dx. This is distinct from the chain rule used on the y to the fourth term alone.