The equation of a curve is given by $xy - 2y^{2} + 4x^{2} = 66$.

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

$\frac{d}{dx}(xy - 2y^2 + 4x^2) = \frac{d}{dx}(66)$
$y + x\frac{dy}{dx} - 4y\frac{dy}{dx} + 8x = 0$

2. Rearrange the equation to express $\frac{dy}{dx}$ in terms of $x$ and $y$.

$\frac{dy}{dx}(x - 4y) + y + 8x = 0$
$\frac{dy}{dx} = \frac{-(8x + y)}{x - 4y}$

3. Apply the condition for the tangent to be parallel to the $y$-axis.

For a tangent parallel to the $y$-axis (vertical tangent), the gradient $\frac{dy}{dx}$ is undefined, which means the denominator must be zero:
$x - 4y = 0 \implies x = 4y$

4. Substitute the relationship into the curve equation to solve for $y$.

Substitute $x = 4y$ into the original equation $xy - 2y^2 + 4x^2 = 66$:
$(4y)(y) - 2y^2 + 4(4y)^2 = 66$
$4y^2 - 2y^2 + 64y^2 = 66$
$66y^2 = 66$
$y^2 = 1 \implies y = 1 \text{ or } y = -1$

5. Find the corresponding $x$-coordinates to determine the exact points.

When $y = 1$:
$x = 4(1) = 4$
When $y = -1$:
$x = 4(-1) = -4$
The exact coordinates are $(4, 1)$ and $(-4, -1)$.

1. Set up the quadratic equation for the intersection of the curve and a line parallel to the $x$-axis.

Let the equation of a line parallel to the $x$-axis be $y = k$, where $k$ is a constant.
Substitute $y = k$ into the equation of the curve:
$x(k) - 2(k)^2 + 4x^2 = 66$
Rearranging this into a quadratic equation in $x$:
$4x^2 + kx - (2k^2 + 66) = 0$

2. Formulate and simplify the discriminant of the quadratic equation.

To determine the number of real roots (which corresponds to the number of intersection points), we examine the discriminant $\Delta$ of the quadratic equation in $x$:
$\Delta = b^2 - 4ac$
$\Delta = (k)^2 - 4(4)[-(2k^2 + 66)]$
$\Delta = k^2 + 16(2k^2 + 66)$
$\Delta = k^2 + 32k^2 + 1056$
$\Delta = 33k^2 + 1056$

3. Analyze the discriminant and conclude the proof.

Since the square of any real number is non-negative, $k^2 \ge 0$ for all real values of $k$.
It follows that $33k^2 \ge 0$.
Thus, $\Delta = 33k^2 + 1056 \ge 1056 > 0$.
Because the discriminant is strictly positive for any real constant $k$, the quadratic equation will always have two distinct real roots for $x$.
Therefore, every line parallel to the $x$-axis cuts the curve at exactly two distinct points. (Shown)