To find the gradient of the curve, we need to differentiate \(y = \frac{e^{2x}}{1 + e^{2x}}\) with respect to \(x\).

Using the quotient rule, \(\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\), where \(u = e^{2x}\) and \(v = 1 + e^{2x}\).

First, find \(\frac{du}{dx} = 2e^{2x}\) and \(\frac{dv}{dx} = 2e^{2x}\).

Substitute into the quotient rule:

\(\frac{dy}{dx} = \frac{(1 + e^{2x})(2e^{2x}) - e^{2x}(2e^{2x})}{(1 + e^{2x})^2}\)

Simplify the numerator:

\(= \frac{2e^{2x} + 2e^{4x} - 2e^{4x}}{(1 + e^{2x})^2}\)

\(= \frac{2e^{2x}}{(1 + e^{2x})^2}\)

Now, substitute \(x = \ln 3\):

\(e^{2x} = e^{2 \ln 3} = 3^2 = 9\).

Substitute into the derivative:

\(\frac{dy}{dx} = \frac{2 \times 9}{(1 + 9)^2} = \frac{18}{100} = \frac{9}{50}\)

Thus, the gradient at \(x = \ln 3\) is \(\frac{9}{50}\).

To find the stationary point, we need to find the derivative of \(y = \frac{\ln x}{x^3}\) and set it to zero.

Using the quotient rule, \(\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\), where \(u = \ln x\) and \(v = x^3\).

\(\frac{du}{dx} = \frac{1}{x}\) and \(\frac{dv}{dx} = 3x^2\).

Thus, \(\frac{dy}{dx} = \frac{x^3 \cdot \frac{1}{x} - \ln x \cdot 3x^2}{x^6} = \frac{x^2 - 3x^2 \ln x}{x^6} = \frac{1 - 3 \ln x}{x^4}\).

Set \(\frac{dy}{dx} = 0\):

\(\frac{1 - 3 \ln x}{x^4} = 0\).

This implies \(1 - 3 \ln x = 0\).

Solving for \(x\), we get \(3 \ln x = 1\).

\(\ln x = \frac{1}{3}\).

Therefore, \(x = e^{1/3}\), which is approximately 1.40.

To find the stationary points, we need to find where the derivative \(\frac{dy}{dx} = 0\).

Using the product rule, the derivative of \(y = e^{-3x} \tan x\) is:

\(\frac{dy}{dx} = e^{-3x} \cdot \frac{d}{dx}(\tan x) + \tan x \cdot \frac{d}{dx}(e^{-3x})\)

\(\frac{dy}{dx} = e^{-3x} \sec^2 x - 3e^{-3x} \tan x\)

Setting \(\frac{dy}{dx} = 0\):

\(e^{-3x} \sec^2 x - 3e^{-3x} \tan x = 0\)

\(\sec^2 x = 3 \tan x\)

\(\frac{1}{\cos^2 x} = 3 \frac{\sin x}{\cos x}\)

\(\cos x = 3 \sin x\)

\(\tan x = \frac{1}{3}\)

Solving \(\tan x = \frac{1}{3}\) within the interval \(-\frac{1}{2}\pi < x < \frac{1}{2}\pi\), we find:

\(x = \arctan(\frac{1}{3}) \approx 0.365\)

Considering the periodicity of the tangent function, the next solution in the interval is:

\(x = \arctan(\frac{1}{3}) + \pi \approx 1.206\)

Thus, the x-coordinates of the stationary points are \(0.365\) and \(1.206\).

To find the stationary point, we need to find the derivative of \(y = \frac{e^x}{\cos x}\) and set it to zero.

Using the quotient rule, the derivative is:

\(\frac{d}{dx} \left( \frac{e^x}{\cos x} \right) = \frac{e^x \cos x + e^x \sin x}{\cos^2 x}\)

Set the derivative to zero:

\(e^x \cos x + e^x \sin x = 0\)

Factor out \(e^x\):

\(e^x (\cos x + \sin x) = 0\)

Since \(e^x \neq 0\), we have:

\(\cos x + \sin x = 0\)

This simplifies to:

\(\tan x = -1\)

The solution for \(x\) in the interval \(-\frac{1}{2}\pi < x < \frac{1}{2}\pi\) is:

\(x = -\frac{1}{4}\pi\)

(i) To find the stationary point, we need to differentiate \(y = e^{-x} \sin x\) and set the derivative to zero. Using the product rule, \(\frac{dy}{dx} = e^{-x} \cos x - e^{-x} \sin x\). Setting \(\frac{dy}{dx} = 0\), we have:

\(e^{-x} (\cos x - \sin x) = 0\)

Since \(e^{-x} \neq 0\), we solve \(\cos x = \sin x\).

This gives \(x = \frac{\pi}{4}\) within the interval \(0 \leq x \leq \pi\).

(ii) To determine the nature of the stationary point, we examine the second derivative or the sign change of the first derivative around \(x = \frac{\pi}{4}\). The second derivative test or analyzing the behavior of \(\frac{dy}{dx}\) shows that it is a maximum point.