Q.If $x = y^3 + 2y$, find $\dfrac{dy}{dx}$, and evaluate it at the point where $y = 1$.

The relation gives $x$ in terms of $y$, so find $\tfrac{dx}{dy}$ and take its reciprocal.

The relation is written as $x$ in terms of $y$, which is exactly when the inverse-function rule (§4) is convenient — differentiate with respect to $y$, then reciprocate.

Differentiate $x$ with respect to $y$.

Apply the inverse-function rule $\dfrac{dy}{dx} = \dfrac{1}{\,dx/dy\,}$:

Evaluate at $y = 1$. $\dfrac{dy}{dx}\Big|_{y=1} = \dfrac{1}{3(1)^2 + 2} = \dfrac{1}{5}$. …

The answer is legitimately left in terms of $y$, since the curve is naturally parametrised by $y$. Solving the cubic $x=y^3+2y$ for $y$ explicitly wo …

The main error is differentiating $y^3+2y$ with respect to $x$ instead of $y$, or forgetting to reciprocate and reporting $3y^2+2$ as $\tfrac{dy}{dx}$ (that is actually $\tfrac{dx}{dy}$). Read carefully which variable i …