Given: \(y = (x + \sqrt{x^2 - 1})^2\)

Differentiate \(y\) with respect to \(x\):

\(\frac{dy}{dx} = 2(x + \sqrt{x^2 - 1}) \cdot \frac{d}{dx}(x + \sqrt{x^2 - 1})\)

Now, differentiate \(x + \sqrt{x^2 - 1}\) with respect to \(x\):

\(\frac{d}{dx}(x + \sqrt{x^2 - 1}) = 1 + \frac{1}{2\sqrt{x^2 - 1}} \cdot 2x = 1 + \frac{x}{\sqrt{x^2 - 1}} = \frac{\sqrt{x^2 - 1} + x}{\sqrt{x^2 - 1}}\)

Substitute this back into the expression for \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = 2(x + \sqrt{x^2 - 1}) \cdot \frac{x + \sqrt{x^2 - 1}}{\sqrt{x^2 - 1}} = \frac{2(x + \sqrt{x^2 - 1})^2}{\sqrt{x^2 - 1}}\)

Since \(y = (x + \sqrt{x^2 - 1})^2\), we can write:

\(\frac{dy}{dx} = \frac{2y}{\sqrt{x^2 - 1}}\)

Now, square both sides:

\((\frac{dy}{dx})^2 = \frac{4y^2}{x^2 - 1}\)

Multiply both sides by \((x^2 - 1)\):

\((x^2 - 1)(\frac{dy}{dx})^2 = 4y^2\)