Implicit differentiation of a trigonometric-algebraic equation

Differentiate the left side using the chain rule: $\frac{d}{dx}[\cos(x-y)] = -\sin(x-y)\cdot(1 - y')$, since $\frac{d}{dx}(x-y) = 1 - y'$. Differentiate the right side using the product rule: $\frac{d}{dx}[xy^2] = y^2 + x\cdot 2y y' = y^2 + 2xy y'$. Setting these equal gives $-\sin(x-y)(1-y') = y^2 + 2xyy'$.

Expand the left side: $-\sin(x-y) + \sin(x-y)y' = y^2 + 2xyy'$. Move all $y'$ terms to one side: $\sin(x-y)y' - 2xyy' = y^2 + \sin(x-y)$. Factor out $y'$ and divide: $y' = \dfrac{y^2 + \sin(x-y)}{\sin(x-y) - 2xy}$.

The right side of the equation was x times y squared, a product of two functions of x. To differentiate it, the product rule was applied, giving y squared plus x times two y times dy dx. The chain rule was also needed inside this step to differentiate y squared with respect to x. So the primary rule governing that term's structure is the product rule.