A curve $C$ has equation $3x^2 - 2xy + 5y^2 = 14.$

1. Apply implicit differentiation and the product rule to differentiate the equation with respect to $x$.

$\frac{d}{dx}(3x^2) - \frac{d}{dx}(2xy) + \frac{d}{dx}(5y^2) = \frac{d}{dx}(14)$
$6x - \left( 2y + 2x \frac{dy}{dx} \right) + 10y \frac{dy}{dx} = 0$
$6x - 2y - 2x \frac{dy}{dx} + 10y \frac{dy}{dx} = 0$

2. Group the terms containing $\frac{dy}{dx}$ and factorize.

$6x - 2y = 2x \frac{dy}{dx} - 10y \frac{dy}{dx}$
$6x - 2y = (2x - 10y)\frac{dy}{dx}$

3. Isolate $\frac{dy}{dx}$ and simplify by factoring out a common factor of 2 to obtain the final expression.

$\frac{dy}{dx} = \frac{6x - 2y}{2x - 10y}$
$\frac{dy}{dx} = \frac{2(3x - y)}{2(x - 5y)}$
$\frac{dy}{dx} = \frac{3x - y}{x - 5y}$

1. Set the denominator of $\frac{dy}{dx}$ to 0 to establish the condition for a vertical tangent.

For a tangent to be parallel to the y-axis, the gradient $\frac{dy}{dx}$ is undefined. This occurs when the denominator is zero:
$x - 5y = 0$
$y = \frac{x}{5}$

2. Substitute the expression for $y$ into the original equation of the curve.

$3x^2 - 2x\left(\frac{x}{5}\right) + 5\left(\frac{x}{5}\right)^2 = 14$
$3x^2 - \frac{2x^2}{5} + 5\left(\frac{x^2}{25}\right) = 14$
$3x^2 - \frac{2x^2}{5} + \frac{x^2}{5} = 14$

3. Solve the resulting quadratic equation for the exact $x$-coordinates.

Multiply the entire equation by 5 to clear the denominators:
$15x^2 - 2x^2 + x^2 = 70$
$14x^2 = 70$
$x^2 = 5$
$x = \pm\sqrt{5}$

1. Find the possible $x$-coordinates on the curve when $y = 1$.

Substitute $y = 1$ into the equation of the curve $C$:
$3x^2 - 2x(1) + 5(1)^2 = 14$
$3x^2 - 2x - 9 = 0$
Using the quadratic formula to solve for $x$:
$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(3)(-9)}}{2(3)}$
$x = \frac{2 \pm \sqrt{112}}{6} = \frac{1 \pm 2\sqrt{7}}{3}$

2. Relate the rates of change using the chain rule to form an expression for $\frac{dx}{dt}$.

Using the chain rule, $\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$.
Given $\frac{dy}{dt} = -7$ and from part (i) $\frac{dy}{dx} = \frac{3x - y}{x - 5y}$, substitute $y = 1$:
$-7 = \left( \frac{3x - 1}{x - 5} \right) \frac{dx}{dt}$
$\frac{dx}{dt} = \frac{-7(x - 5)}{3x - 1}$

3. Determine which value of $x$ gives $\frac{dx}{dt} > 0$ and calculate the exact rate of change.

For $x = \frac{1 - 2\sqrt{7}}{3} \approx -1.43$:
$x - 5 < 0$ and $3x - 1 < 0$.
Thus, $\frac{dx}{dt} = \frac{-7(-)}{-} < 0$, which contradicts the condition that $x$ increases.

For $x = \frac{1 + 2\sqrt{7}}{3} \approx 2.10$:
$x - 5 < 0$ and $3x - 1 > 0$.
Thus, $\frac{dx}{dt} = \frac{-7(-)}{+} > 0$, which satisfies the condition.

Substitute the accepted $x = \frac{1 + 2\sqrt{7}}{3}$ to find the exact rate:
$\frac{dx}{dt} = \frac{-7\left( \frac{1 + 2\sqrt{7}}{3} - 5 \right)}{3\left( \frac{1 + 2\sqrt{7}}{3} \right) - 1} = \frac{-7\left( \frac{2\sqrt{7} - 14}{3} \right)}{2\sqrt{7}}$
$\frac{dx}{dt} = \frac{98 - 14\sqrt{7}}{6\sqrt{7}} = \frac{49 - 7\sqrt{7}}{3\sqrt{7}}$
Rationalising the denominator:
$\frac{dx}{dt} = \frac{(49 - 7\sqrt{7})\sqrt{7}}{21} = \frac{49\sqrt{7} - 49}{21} = \frac{7(\sqrt{7} - 1)}{3}$