Collect the derivative terms in implicit relations

On a differentiable local branch $y=y(x)$, differentiating gives $2x+3y^2y^{\prime}=2xyy^{\prime}+y^2$. The right side needs both product terms, including the derivative of $x$.

$(3y^2-2xy)y^{\prime}=y^2-2x$, so $y^{\prime}=(y^2-2x)/(3y^2-2xy)$ where $3y^2-2xy\ne0$. This is the source answer. At a zero denominator, this division does not classify the point or prove a finite derivative exists.

The Student copy leaves this working blank; the Teacher supplies $-\sin x=y(-\sin y)y^{\prime}+\cos y\,y^{\prime}$. Both factors in $y\cos y$ depend on $x$.

Rearrange to $(y\sin y-\cos y)y^{\prime}=\sin x$. Hence $y^{\prime}=\sin x/(y\sin y-\cos y)$ where the denominator is nonzero, as printed in the Teacher copy. Angles are in radians.