1. **Problem Statement:** Given the equation $$y^2 + 2ax - a^2 = 0,$$ show that $$\left(\frac{dy}{dx}\right)^2 + \frac{2x}{y} \frac{dy}{dx} = 1.$$\n\n2. **Step 1: Differentiate the given equation implicitly with respect to $x$.**\nThe equation is $$y^2 + 2ax - a^2 = 0.$$\nDifferentiating term-by-term:\n$$\frac{d}{dx}(y^2) + \frac{d}{dx}(2ax) - \frac{d}{dx}(a^2) = 0.$$\nSince $a$ is a constant, $\frac{d}{dx}(a^2) = 0$.\n\n3. **Apply the chain rule and derivative rules:**\n$$2y \frac{dy}{dx} + 2a = 0.$$\n\n4. **Solve for $\frac{dy}{dx}$:**\n$$2y \frac{dy}{dx} = -2a \implies \frac{dy}{dx} = -\frac{a}{y}.$$\n\n5. **Substitute $\frac{dy}{dx}$ into the expression to prove:**\nWe want to show:\n$$\left(\frac{dy}{dx}\right)^2 + \frac{2x}{y} \frac{dy}{dx} = 1.$$\nSubstitute $\frac{dy}{dx} = -\frac{a}{y}$:\n$$\left(-\frac{a}{y}\right)^2 + \frac{2x}{y} \left(-\frac{a}{y}\right) = \frac{a^2}{y^2} - \frac{2ax}{y^2} = \frac{a^2 - 2ax}{y^2}.$$\n\n6. **Use the original equation to simplify:**\nFrom the original equation, rearranged:\n$$y^2 = a^2 - 2ax.$$\nSubstitute into the numerator:\n$$\frac{a^2 - 2ax}{y^2} = \frac{y^2}{y^2} = 1.$$\n\n7. **Conclusion:**\nThus,\n$$\left(\frac{dy}{dx}\right)^2 + \frac{2x}{y} \frac{dy}{dx} = 1,$$\nas required.