已知 $y = \sqrt{1 + \tan^{-1} 2x}$，其中 $-\frac{1}{2} < x < \frac{1}{2}$。证明 $y \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + \frac{8x}{(1 + 4x^2)^2} = 0$ 由此求 $y$ 的麦克劳林级数，展开至包含 $x^3$ 的项。

1. Square the given equation and differentiate implicitly with respect to $x$.

$y = \sqrt{1 + \tan^{-1} 2x}$
$y^2 = 1 + \tan^{-1} 2x$

Differentiating both sides with respect to $x$:
$2y \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{1 + (2x)^2} \cdot 2$
$2y \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2}{1 + 4x^2}$
$y \frac{\mathrm{d}y}{\mathrm{d}x} = (1 + 4x^2)^{-1}$

2. Differentiate a second time using the product and chain rules to establish the given equation.

Differentiating with respect to $x$ again:
$y \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \frac{\mathrm{d}y}{\mathrm{d}x} \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right) = -1(1 + 4x^2)^{-2} (8x)$
$y \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = -\frac{8x}{(1 + 4x^2)^2}$
$y \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + \frac{8x}{(1 + 4x^2)^2} = 0 \quad \text{(shown)}$

3. Evaluate $y$, $\frac{\mathrm{d}y}{\mathrm{d}x}$, and $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$ at $x = 0$.

When $x = 0$:
$y^2 = 1 + \tan^{-1}(0) = 1 \implies y = 1 \quad (\text{since } y > 0)$
$1 \cdot \frac{\mathrm{d}y}{\mathrm{d}x} = (1 + 0)^{-1} \implies \frac{\mathrm{d}y}{\mathrm{d}x} = 1$
$1 \cdot \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + (1)^2 + 0 = 0 \implies \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -1$

4. Differentiate the second derivative equation implicitly to find an expression for the third derivative.

$\frac{\mathrm{d}}{\mathrm{d}x} \left( y \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + 8x(1 + 4x^2)^{-2} \right) = 0$
$y \frac{\mathrm{d}^3y}{\mathrm{d}x^3} + \frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 2\frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 8(1 + 4x^2)^{-2} + 8x(-2)(1 + 4x^2)^{-3}(8x) = 0$
$y \frac{\mathrm{d}^3y}{\mathrm{d}x^3} + 3\frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 8(1 + 4x^2)^{-2} - 128x^2(1 + 4x^2)^{-3} = 0$

5. Evaluate $\frac{\mathrm{d}^3y}{\mathrm{d}x^3}$ at $x = 0$.

Substituting $x = 0$, $y = 1$, $\frac{\mathrm{d}y}{\mathrm{d}x} = 1$, $\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -1$:
$1 \cdot \frac{\mathrm{d}^3y}{\mathrm{d}x^3} + 3(1)(-1) + 8(1)^{-2} - 0 = 0$
$\frac{\mathrm{d}^3y}{\mathrm{d}x^3} - 3 + 8 = 0 \implies \frac{\mathrm{d}^3y}{\mathrm{d}x^3} = -5$

6. Construct the Maclaurin series up to the term in $x^3$.

$y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \dots$
$y = 1 + (1)x + \frac{-1}{2!}x^2 + \frac{-5}{3!}x^3$
$y = 1 + x - \frac{1}{2}x^2 - \frac{5}{6}x^3$