已知 $y = \sqrt{1+\tan x}$。

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

$y = \sqrt{1+\tan x}$
$y^2 = 1 + \tan x$
Differentiating implicitly with respect to $x$:
$2y\frac{dy}{dx} = \sec^2 x$

2. Use the appropriate trigonometric identity to substitute $\sec^2 x$ in terms of $y$ to show the first result.

Using the trigonometric identity $\sec^2 x = 1 + \tan^2 x$:
$2y\frac{dy}{dx} = 1 + \tan^2 x$
Since $y^2 = 1 + \tan x$, we have $\tan x = y^2 - 1$. Substituting this in:
$2y\frac{dy}{dx} = 1 + (y^2 - 1)^2$

3. Differentiate the first result implicitly with respect to $x$ to find an expression involving the second derivative.

Applying the product rule on the left-hand side and the chain rule on the right-hand side:
$\frac{d}{dx}\left(2y\frac{dy}{dx}\right) = \frac{d}{dx}\left(1 + (y^2 - 1)^2\right)$
$2y\frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right)\left(\frac{dy}{dx}\right) = 2(y^2 - 1) \cdot \frac{d}{dx}(y^2 - 1)$
$2y\frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right)^2 = 2(y^2 - 1)\left(2y\frac{dy}{dx}\right)$

4. Simplify the resulting equation to establish the second result.

Dividing the entire equation by 2:
$y\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 2y(y^2 - 1)\left(\frac{dy}{dx}\right)$

1. Evaluate the value of $y$ at $x = 0$.

Given $y = \sqrt{1+\tan x}$.
When $x = 0$,
$y = \sqrt{1+\tan 0} = \sqrt{1+0} = 1$

2. Evaluate the first derivative $\dfrac{dy}{dx}$ at $x = 0$ using the first given identity.

Substitute $x = 0$ and $y = 1$ into $2y\,\dfrac{dy}{dx} = 1 + (y^{2}-1)^{2}$:
$2(1)\left.\dfrac{dy}{dx}\right|_{x=0} = 1 + (1^{2}-1)^{2}$
$2\left.\dfrac{dy}{dx}\right|_{x=0} = 1 + 0$
$\left.\dfrac{dy}{dx}\right|_{x=0} = \dfrac{1}{2}$

3. Evaluate the second derivative $\dfrac{d^{2}y}{dx^{2}}$ at $x = 0$ using the second given identity.

Substitute $x = 0$, $y = 1$, and $\dfrac{dy}{dx} = \dfrac{1}{2}$ into $y\,\dfrac{d^{2}y}{dx^{2}} + \left(\dfrac{dy}{dx}\right)^{2} = 2y(y^{2}-1)\left(\dfrac{dy}{dx}\right)$:
$(1)\left.\dfrac{d^{2}y}{dx^{2}}\right|_{x=0} + \left(\dfrac{1}{2}\right)^{2} = 2(1)(1^{2}-1)\left(\dfrac{1}{2}\right)$
$\left.\dfrac{d^{2}y}{dx^{2}}\right|_{x=0} + \dfrac{1}{4} = 0$
$\left.\dfrac{d^{2}y}{dx^{2}}\right|_{x=0} = -\dfrac{1}{4}$

4. Form the Maclaurin series expansion up to the first three terms.

The Maclaurin series expansion is given by:
$y \approx y(0) + x\left(\dfrac{dy}{dx}\right)_{x=0} + \dfrac{x^{2}}{2!}\left(\dfrac{d^{2}y}{dx^{2}}\right)_{x=0}$
$y \approx 1 + x\left(\dfrac{1}{2}\right) + \dfrac{x^{2}}{2}\left(-\dfrac{1}{4}\right)$
$y \approx 1 + \dfrac{1}{2}x - \dfrac{1}{8}x^{2}$