曲线 C 的方程为 $2x - y^{2} = (x + y)^{2}$。

1. Differentiate the equation of the curve implicitly with respect to $x$.

$2x - y^2 = (x + y)^2$

Differentiating both sides with respect to $x$:
$2 - 2y \frac{dy}{dx} = 2(x + y)\left(1 + \frac{dy}{dx}\right)$

2. Use the condition for horizontal tangents to find a relationship between $x$ and $y$.

Tangents parallel to the $x$-axis have a gradient of $0$, so $\frac{dy}{dx} = 0$.

Substituting $\frac{dy}{dx} = 0$ into the differentiated equation:
$2 - 2y(0) = 2(x + y)(1 + 0)$
$2 = 2(x + y)$
$1 = x + y$
$x = 1 - y$

3. Substitute the relationship back into the original curve equation to solve for $y$.

Substitute $x = 1 - y$ into $2x - y^2 = (x + y)^2$:
$2(1 - y) - y^2 = (1 - y + y)^2$
$2 - 2y - y^2 = 1^2$
$2 - 2y - y^2 = 1$
$y^2 + 2y - 1 = 0$

Using the quadratic formula:
$y = \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2(1)}$
$y = \frac{-2 \pm \sqrt{8}}{2}$
$y = \frac{-2 \pm 2\sqrt{2}}{2}$
$y = -1 \pm \sqrt{2}$

4. State the equations of the tangents.

Since the tangents are parallel to the $x$-axis, their equations take the form $y = c$.

The equations of the tangents are $y = -1 + \sqrt{2}$ and $y = -1 - \sqrt{2}$.

1. Find the derivative $\frac{dy}{dx}$ using implicit differentiation (from part i).

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

2. Find the equation of the tangent line $l$ at point $A(2, -2)$.

Substitute $x = 2$ and $y = -2$ into $\frac{dy}{dx}$ to find the gradient $m_l$:
$m_l = \frac{1 - 2 - (-2)}{2 + 2(-2)} = \frac{1}{-2} = -\frac{1}{2}$
The equation of the tangent line $l$ is:
$y - (-2) = -\frac{1}{2}(x - 2)$
$y = -\frac{1}{2}x - 1$

3. Determine the equation of the normal to $C$ at the origin $O$.

At the origin $O(0,0)$:
$\frac{dy}{dx} = \frac{1 - 0 - 0}{0 + 0} = \frac{1}{0}$
Since the derivative evaluates to a non-zero number divided by zero, the gradient of the tangent at $O$ is undefined. This means the tangent is the vertical line $x = 0$ (the $y$-axis).
Therefore, the normal to the curve at $O$ is perpendicular to the $y$-axis, which is the horizontal line $y = 0$ (the $x$-axis).

4. Find the coordinates of $B$ where the normal at $O$ meets the tangent $l$.

Substitute $y = 0$ into the equation of $l$ to find the intersection point $B$:
$0 = -\frac{1}{2}x - 1$
$\frac{1}{2}x = -1 \implies x = -2$
Thus, the coordinates of $B$ are $(-2, 0)$.

5. Calculate the area of triangle $OAB$.

The vertices of $\triangle OAB$ are $O(0,0)$, $A(2, -2)$ and $B(-2, 0)$.
Taking $OB$ along the $x$-axis as the base, its length is $2$ units.
The perpendicular height of $\triangle OAB$ is the absolute $y$-coordinate of $A$, which is $|-2| = 2$ units.
$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 2 = 2$