Let $x$ and $y$ be variables such that $\ln(y + 1) = 1 + \tan^{-1}x$.

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

$\frac{1}{y+1}\frac{\text{d}y}{\text{d}x} = \frac{1}{1+x^2}$

2. Rearrange the equation to a linear form to simplify further differentiation.

$(1+x^2)\frac{\text{d}y}{\text{d}x} = y+1$

3. Differentiate implicitly a second time with respect to $x$ using the product rule.

$(1+x^2)\frac{\text{d}^2y}{\text{d}x^2} + 2x\frac{\text{d}y}{\text{d}x} = \frac{\text{d}y}{\text{d}x}$

4. Rearrange the terms to show the required result.

$(1+x^2)\frac{\text{d}^2y}{\text{d}x^2} = \frac{\text{d}y}{\text{d}x} - 2x\frac{\text{d}y}{\text{d}x}

$(1+x^2)\frac{\text{d}^2y}{\text{d}x^2} = (1-2x)\frac{\text{d}y}{\text{d}x}$

1. Find the value of $y$ when $x = 0$.

Substitute $x = 0$ into the original equation $\ln(y+1) = 1 + \tan^{-1}x$:
$\ln(y+1) = 1 + \tan^{-1}(0)$
$\ln(y+1) = 1$
$y+1 = e \implies y = e-1$

2. Find the first derivative $\frac{\text{d}y}{\text{d}x}$ at $x = 0$.

Differentiate $\ln(y+1) = 1 + \tan^{-1}x$ with respect to $x$:
$\frac{1}{y+1} \frac{\text{d}y}{\text{d}x} = \frac{1}{1+x^2}$
Substitute $x = 0$ and $y = e-1$:
$\frac{1}{e} \frac{\text{d}y}{\text{d}x} = 1 \implies \frac{\text{d}y}{\text{d}x} = e$

3. Find the second derivative $\frac{\text{d}^2y}{\text{d}x^2}$ at $x = 0$.

Using the equation given in part (i):
$(1+x^2)\frac{\text{d}^2y}{\text{d}x^2} = (1-2x)\frac{\text{d}y}{\text{d}x}$
Substitute $x = 0$ and $\frac{\text{d}y}{\text{d}x} = e$:
$(1+0)\frac{\text{d}^2y}{\text{d}x^2} = (1-0)(e)$
$\frac{\text{d}^2y}{\text{d}x^2} = e$

4. Find the third derivative $\frac{\text{d}^3y}{\text{d}x^3}$ at $x = 0$ by further differentiation.

Differentiate the equation from part (i) with respect to $x$ using the product rule:
$(1+x^2)\frac{\text{d}^3y}{\text{d}x^3} + 2x\frac{\text{d}^2y}{\text{d}x^2} = (1-2x)\frac{\text{d}^2y}{\text{d}x^2} - 2\frac{\text{d}y}{\text{d}x}$
Substitute $x = 0$, $\frac{\text{d}y}{\text{d}x} = e$, and $\frac{\text{d}^2y}{\text{d}x^2} = e$:
$(1)\frac{\text{d}^3y}{\text{d}x^3} + 0 = (1)(e) - 2(e)$
$\frac{\text{d}^3y}{\text{d}x^3} = -e$

5. Construct the Maclaurin series for $y$.

Using the Maclaurin series formula $y(x) = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \dots$:
$y = (e-1) + ex + \frac{e}{2!}x^2 + \frac{-e}{3!}x^3 + \dots$
$y = e - 1 + ex + \frac{e}{2}x^2 - \frac{e}{6}x^3 + \dots$

1. Find the first two non-zero terms of the Maclaurin series for $y$ by evaluating $y$ and $\frac{\text{d}y}{\text{d}x}$ at $x=0$.

At $x=0$:
$\ln(y+1) = 1 + \tan^{-1}(0) = 1 \implies y+1 = e \implies y = e-1$

Differentiating both sides with respect to $x$ gives:
$\frac{1}{y+1}\frac{\text{d}y}{\text{d}x} = \frac{1}{1+x^2}$
At $x=0$:
$\frac{1}{e}\frac{\text{d}y}{\text{d}x} = 1 \implies \frac{\text{d}y}{\text{d}x} = e$

The first two non-zero terms of the Maclaurin series for $y$ are $(e-1)$ and $ex$.

2. Find the first two non-zero terms of the binomial series expansion of $\frac{1}{a+bx}$.

$\frac{1}{a+bx} = (a+bx)^{-1}$
$= a^{-1}\left(1 + \frac{b}{a}x\right)^{-1}$
$= \frac{1}{a} \left( 1 + (-1)\left(\frac{b}{a}x\right) + \dots \right)$
$= \frac{1}{a} - \frac{b}{a^2}x + \dots$

The first two non-zero terms are $\frac{1}{a}$ and $-\frac{b}{a^2}x$.

3. Equate the corresponding terms to solve for $a$ and $b$.

Equating the constant terms:
$\frac{1}{a} = e - 1 \implies a = \frac{1}{e-1}$

Equating the coefficients of $x$:
$-\frac{b}{a^2} = e$
$b = -e a^2 = -e \left( \frac{1}{e-1} \right)^2 = -\frac{e}{(e-1)^2}$