对 $\frac{\sin^{-1}(2x)}{1 - 4x^2}$（其中 $-\frac{1}{2} < x < \frac{1}{2}$）关于 $x$ 求导，并尽可能将其化简为单个最简分式。 已知 $y^2 = 3\text{e}^{4x} + 4$。通过多次求导，证明 $y\frac{\text{d}^2 y}{\text{d}x^2} + \left(\frac{\text{d}y}{\text{d}x}\right)^2 - 8y^2 = k$ 其中 $k$ 是一个待确定的常数。

1. Identify the functions for the numerator and denominator and find their derivatives.

Let $u = \sin^{-1}(2x)$ and $v = 1 - 4x^2$.
Using the chain rule, the derivative of $u$ is:
$\frac{\text{d}u}{\text{d}x} = \frac{1}{\sqrt{1 - (2x)^2}} \cdot 2 = \frac{2}{\sqrt{1 - 4x^2}}$
The derivative of $v$ is:
$\frac{\text{d}v}{\text{d}x} = -8x$

2. Apply the quotient rule to find the derivative of the given expression and simplify to a single fraction.

Using the quotient rule $\frac{\text{d}}{\text{d}x}\left(\frac{u}{v}\right) = \frac{v\frac{\text{d}u}{\text{d}x} - u\frac{\text{d}v}{\text{d}x}}{v^2}$:
$\frac{\text{d}}{\text{d}x} \left( \frac{\sin^{-1}(2x)}{1 - 4x^2} \right) = \frac{(1 - 4x^2) \left( \frac{2}{\sqrt{1 - 4x^2}} \right) - \sin^{-1}(2x)(-8x)}{(1 - 4x^2)^2}$
Simplifying the numerator:
$= \frac{2\sqrt{1 - 4x^2} + 8x \sin^{-1}(2x)}{(1 - 4x^2)^2}$

1. Differentiate the given equation implicitly with respect to $x$.

$\frac{\text{d}}{\text{d}x}(y^2) = \frac{\text{d}}{\text{d}x}(3\text{e}^{4x} + 4)$
$2y\frac{\text{d}y}{\text{d}x} = 12\text{e}^{4x}$
$y\frac{\text{d}y}{\text{d}x} = 6\text{e}^{4x}$

2. Differentiate implicitly a second time using the product rule.

$\frac{\text{d}}{\text{d}x}\left( y\frac{\text{d}y}{\text{d}x} \right) = \frac{\text{d}}{\text{d}x}(6\text{e}^{4x})$
$y\frac{\text{d}^2 y}{\text{d}x^2} + \frac{\text{d}y}{\text{d}x}\left(\frac{\text{d}y}{\text{d}x}\right) = 24\text{e}^{4x}$
$y\frac{\text{d}^2 y}{\text{d}x^2} + \left(\frac{\text{d}y}{\text{d}x}\right)^2 = 24\text{e}^{4x}$

3. Express the exponential term in terms of $y^2$ using the original equation.

From the original equation, we have:
$3\text{e}^{4x} = y^2 - 4$
Multiply the entire equation by 8 to match the right-hand side of the second derivative:
$24\text{e}^{4x} = 8(y^2 - 4) = 8y^2 - 32$
Substitute this back into the second derivative equation:
$y\frac{\text{d}^2 y}{\text{d}x^2} + \left(\frac{\text{d}y}{\text{d}x}\right)^2 = 8y^2 - 32$

4. Rearrange to the required form and determine the value of $k$.

$y\frac{\text{d}^2 y}{\text{d}x^2} + \left(\frac{\text{d}y}{\text{d}x}\right)^2 - 8y^2 = -32$
Comparing this with the given identity $y\frac{\text{d}^2 y}{\text{d}x^2} + \left(\frac{\text{d}y}{\text{d}x}\right)^2 - 8y^2 = k$, it follows that:
$k = -32$