Implicit differentiation to find dy/dx

Differentiating both sides with respect to $x$ gives $e^y \cdot \frac{dy}{dx} = 2x + 2y \cdot \frac{dy}{dx}$, using the chain rule on the left and on the $y^2$ term on the right. Now collect the terms containing $\frac{dy}{dx}$ on one side: $e^y \frac{dy}{dx} - 2y \frac{dy}{dx} = 2x$.

Factor out $\frac{dy}{dx}$ to get $\frac{dy}{dx}(e^y - 2y) = 2x$, and then divide both sides by $(e^y - 2y)$ to isolate the derivative. This yields $\frac{dy}{dx} = \dfrac{2x}{e^y - 2y}$.

Starting from $\frac{dy}{dx} = \dfrac{2x}{e^y - 2y}$, apply the quotient rule with numerator $u = 2x$ and denominator $v = e^y - 2y$. We have $u' = 2$, and $v' = e^y\frac{dy}{dx} - 2\frac{dy}{dx} = (e^y - 2)\frac{dy}{dx}$, using the chain rule since $y$ depends on $x$.

The quotient rule gives $\dfrac{d^2y}{dx^2} = \dfrac{2(e^y - 2y) - 2x(e^y - 2)\frac{dy}{dx}}{(e^y - 2y)^2}$. To simplify, substitute $\frac{dy}{dx} = \dfrac{2x}{e^y - 2y}$ from part (a) into the numerator: $2(e^y - 2y) - 2x(e^y - 2)\cdot\dfrac{2x}{e^y - 2y} = 2(e^y - 2y) - \dfrac{4x^2(e^y - 2)}{e^y - 2y}$. Combining over a common denominator gives $\dfrac{2(e^y - 2y)^2 - 4x^2(e^y - 2)}{e^y - 2y}$. Finally, dividing by the denominator squared: $\dfrac{d^2y}{dx^2} = \dfrac{2(e^y - 2y)^2 - 4x^2(e^y - 2)}{(e^y - 2y)^3}$.