1. **State the problem:** Find the second derivative $\frac{d^2y}{dx^2}$ given the implicit differentiation result for $\frac{dy}{dx} = \frac{2y - x^2}{y^2 - 2x}$. 2. **Recall the formula:** To find the second derivative, differentiate $\frac{dy}{dx}$ with respect to $x$ again: $$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{2y - x^2}{y^2 - 2x}\right)$$ Use the quotient rule: $$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$$ where $u = 2y - x^2$ and $v = y^2 - 2x$. 3. **Find derivatives of $u$ and $v$:** - $\frac{du}{dx} = 2 \frac{dy}{dx} - 2x$ - $\frac{dv}{dx} = 2y \frac{dy}{dx} - 2$ 4. **Substitute $\frac{dy}{dx}$ into $\frac{du}{dx}$ and $\frac{dv}{dx}$:** $$\frac{dy}{dx} = \frac{2y - x^2}{y^2 - 2x}$$ 5. **Apply the quotient rule:** $$\frac{d^2y}{dx^2} = \frac{(y^2 - 2x)(2 \frac{dy}{dx} - 2x) - (2y - x^2)(2y \frac{dy}{dx} - 2)}{(y^2 - 2x)^2}$$ 6. **This expression gives the second derivative implicitly in terms of $x$, $y$, and $\frac{dy}{dx}$.** **Final answer:** $$\boxed{\frac{d^2y}{dx^2} = \frac{(y^2 - 2x)(2 \frac{dy}{dx} - 2x) - (2y - x^2)(2y \frac{dy}{dx} - 2)}{(y^2 - 2x)^2}}$$