A curve $C$ has equation $y = \frac{e^{kx}}{e^{kx} - e^{-kx}}$, where $x \ne 0$ and $k \in \mathbb{R},\ k \ne 0$.

1. Differentiate the equation of the curve using the quotient rule and simplify.

Using the quotient rule:
$\frac{dy}{dx} = \frac{\left[\frac{d}{dx}(e^{kx})\right](e^{kx} - e^{-kx}) - (e^{kx})\left[\frac{d}{dx}(e^{kx} - e^{-kx})\right]}{(e^{kx} - e^{-kx})^2}$
$\frac{dy}{dx} = \frac{(k e^{kx})(e^{kx} - e^{-kx}) - (e^{kx})(k e^{kx} + k e^{-kx})}{(e^{kx} - e^{-kx})^2}$
Expanding the terms in the numerator:
$\frac{dy}{dx} = \frac{k e^{2kx} - k e^0 - (k e^{2kx} + k e^0)}{(e^{kx} - e^{-kx})^2}$
$\frac{dy}{dx} = \frac{-2k}{(e^{kx} - e^{-kx})^2}$

2. State the condition for stationary points and deduce that none exist.

For stationary points, we require $\frac{dy}{dx} = 0$.
Since $k \neq 0$, the numerator $-2k \neq 0$.
Therefore, $\frac{dy}{dx} \neq 0$ for all valid $x$, demonstrating that $C$ has no stationary points.

1. Find the $x$-coordinate of the point where $y = 2$.

$\frac{e^{kx}}{e^{kx} - e^{-kx}} = 2$
Multiply the numerator and denominator by $e^{kx}$ to simplify the equation:
$\frac{e^{2kx}}{e^{2kx} - 1} = 2$
$2e^{2kx} - 2 = e^{2kx} \implies e^{2kx} = 2$
$2kx = \ln 2 \implies x = \frac{\ln 2}{2k}$

2. Find the gradient function $\frac{dy}{dx}$ (typically derived in part (i)).

Applying the quotient rule to $y = \frac{e^{2kx}}{e^{2kx} - 1}$:
$\frac{dy}{dx} = \frac{(e^{2kx} - 1)(2k e^{2kx}) - (e^{2kx})(2k e^{2kx})}{(e^{2kx} - 1)^2}$
$\frac{dy}{dx} = \frac{2k e^{4kx} - 2k e^{2kx} - 2k e^{4kx}}{(e^{2kx} - 1)^2} = \frac{-2k e^{2kx}}{(e^{2kx} - 1)^2}$

3. Evaluate the gradient of the tangent at $e^{2kx} = 2$.

Substitute $e^{2kx} = 2$ directly into the derivative to find the gradient $m$:
$m = \frac{-2k(2)}{(2 - 1)^2} = -4k$

4. Form the equation of the tangent.

Using the point-slope form $y - y_1 = m(x - x_1)$ with the point $\left(\frac{\ln 2}{2k}, 2\right)$:
$y - 2 = -4k\left(x - \frac{\ln 2}{2k}\right)$
$y - 2 = -4kx + \frac{4k \ln 2}{2k}$
$y - 2 = -4kx + 2\ln 2$
$y = -4kx + 2\ln 2 + 2$

1. State the gradient of the tangent when $k = 1$.

From part (ii), the gradient of the tangent at $y = 2$ is determined to be $m = -4k$.
Substituting $k = 1$:
$m = -4(1) = -4$

2. Calculate the acute angle between the tangent and the y-axis.

Let $\alpha$ be the acute angle the tangent makes with the x-axis.
$\tan \alpha = |-4| = 4$
$\alpha = \tan^{-1}(4) \approx 75.964^\circ$
Since the x-axis and y-axis are perpendicular, the acute angle between the tangent and the y-axis is:
$90^\circ - 75.964^\circ = 14.036^\circ \approx 14.0^\circ$