已知 $y = \sqrt{1 + \tan^{-1} 2x}$，其中 $-\tfrac{1}{2} < x < \tfrac{1}{2}$。

1. Square both sides of the given equation to avoid differentiating a square root.

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

2. Differentiate implicitly with respect to $x$ for the first time.

$2y \frac{dy}{dx} = \frac{1}{1 + (2x)^2} \cdot 2$
$2y \frac{dy}{dx} = \frac{2}{1 + 4x^2}$

3. Simplify the first derivative equation and write it in index form for the second differentiation.

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

4. Differentiate implicitly a second time using the product rule and chain rule.

$y \frac{d^2 y}{dx^2} + \frac{dy}{dx} \left(\frac{dy}{dx}\right) = -1(1 + 4x^2)^{-2} \cdot (8x)$
$y \frac{d^2 y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = -\frac{8x}{(1 + 4x^2)^2}$

5. Rearrange the terms to show the required result.

$y \frac{d^2 y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + \frac{8x}{(1 + 4x^2)^2} = 0$

1. Evaluate $y$ and $\frac{dy}{dx}$ at $x=0$.

When $x = 0$, $y = \sqrt{1 + \tan^{-1}(0)} = 1$.

Differentiating $y = (1 + \tan^{-1} 2x)^{\frac{1}{2}}$ with respect to $x$:
$\frac{dy}{dx} = \frac{1}{2}(1 + \tan^{-1} 2x)^{-\frac{1}{2}} \left( \frac{2}{1+4x^2} \right)$
Substitute $x=0$:
$\frac{dy}{dx} = \frac{1}{2}(1)^{-\frac{1}{2}} \left(\frac{2}{1}\right) = 1$

2. Evaluate $\frac{d^2y}{dx^2}$ at $x=0$ using the result from part (a).

Using the equation given in part (a):
$y\frac{d^2 y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + \frac{8x}{(1+4x^2)^2} = 0$
Substitute $x=0$, $y=1$, and $\frac{dy}{dx}=1$:
$(1)\frac{d^2y}{dx^2} + (1)^2 + 0 = 0$
$\frac{d^2y}{dx^2} = -1$

3. Differentiate the identity from part (a) to find the expression for $\frac{d^3y}{dx^3}$.

Differentiating $y\frac{d^2 y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + 8x(1+4x^2)^{-2} = 0$ with respect to $x$ using the product and chain rules:
$\left( \frac{dy}{dx}\frac{d^2 y}{dx^2} + y\frac{d^3 y}{dx^3} \right) + 2\frac{dy}{dx}\frac{d^2 y}{dx^2} + \left[ 8(1+4x^2)^{-2} + 8x(-2)(1+4x^2)^{-3}(8x) \right] = 0$
$y\frac{d^3 y}{dx^3} + 3\frac{dy}{dx}\frac{d^2 y}{dx^2} + \frac{8}{(1+4x^2)^2} - \frac{128x^2}{(1+4x^2)^3} = 0$

4. Evaluate $\frac{d^3y}{dx^3}$ at $x=0$.

Substitute $x=0$, $y=1$, $\frac{dy}{dx}=1$, and $\frac{d^2y}{dx^2}=-1$ into the differentiated equation:
$(1)\frac{d^3 y}{dx^3} + 3(1)(-1) + \frac{8}{1^2} - 0 = 0$
$\frac{d^3 y}{dx^3} - 3 + 8 = 0$
$\frac{d^3 y}{dx^3} = -5$

5. Formulate the Maclaurin series up to the term in $x^3$.

The Maclaurin series expansion is given by:
$y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \ldots$
Substitute the values found:
$y = 1 + (1)x + \frac{-1}{2}x^2 + \frac{-5}{6}x^3 + \ldots$
$y = 1 + x - \frac{1}{2}x^2 - \frac{5}{6}x^3$