The curve $C$ has equation $2x - y^2 = (x + y)^2$.

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

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

2. Apply the condition for tangents parallel to the $x$-axis.

For tangents parallel to the $x$-axis, $\frac{dy}{dx} = 0$.
$2 - 2y(0) = 2(x+y)(1 + 0)$
$2 = 2(x+y)$
$x + y = 1$
$x = 1 - y$

3. Substitute the relationship between $x$ and $y$ into the original curve equation.

$2(1-y) - y^2 = (1-y+y)^2$
$2 - 2y - y^2 = 1^2$
$y^2 + 2y - 1 = 0$

4. Solve the resulting quadratic equation for $y$ to find the equations of the tangents.

$y = \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2}$
$y = \frac{-2 \pm \sqrt{8}}{2}$
$y = -1 \pm \sqrt{2}$
Since the tangents are horizontal, their equations are $y = -1 - \sqrt{2}$ and $y = -1 + \sqrt{2}$.

1. Find the derivative $\frac{dy}{dx}$ using implicit differentiation.

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

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

Dividing both sides by $2$ and expanding:
$1 - y\frac{dy}{dx} = x + x\frac{dy}{dx} + y + y\frac{dy}{dx}$

Rearranging to make $\frac{dy}{dx}$ the subject:
$1 - x - y = (x + 2y)\frac{dy}{dx}$
$\frac{dy}{dx} = \frac{1 - x - y}{x + 2y}$

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

Substitute $x = 2$ and $y = -2$ into the derivative to find the gradient of the tangent at $A$:
$\frac{dy}{dx} = \frac{1 - 2 - (-2)}{2 + 2(-2)} = \frac{1}{-2} = -\frac{1}{2}$

Using the point-slope form, the equation of the tangent $l$ is:
$y - (-2) = -\frac{1}{2}(x - 2)$
$y + 2 = -\frac{1}{2}x + 1$
$y = -\frac{1}{2}x - 1$

3. Find the equation of the normal at $O$ and its intersection $B$ with $l$.

At the origin $O(0,0)$:
$\frac{dy}{dx} = \frac{1 - 0 - 0}{0 + 2(0)} = \frac{1}{0}$ (undefined)

Since the gradient is undefined, the tangent at $O$ is the vertical line $x = 0$. The normal is perpendicular to the tangent, making it a horizontal line passing through $(0,0)$.
Equation of normal at $O$: $y = 0$

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

4. Calculate the area of triangle $OAB$.

The vertices of the triangle are $O(0,0)$, $A(2,-2)$, and $B(-2,0)$.

Taking $OB$ (which lies on the $x$-axis) as the base, its length is $2$ units.
The perpendicular height of triangle $OAB$ is the vertical distance from $A$ to the $x$-axis, which is $|-2| = 2$ units.

Area of $\triangle OAB = \frac{1}{2} \times \text{base} \times \text{height}$
Area = $\frac{1}{2} \times 2 \times 2 = 2$ square units.