To find the gradient of the curve, we need to differentiate \(y = \frac{\sin x}{1 + \cos x}\) 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 \(u = \sin x\) and \(v = 1 + \cos x\).

Then, \(\frac{du}{dx} = \cos x\) and \(\frac{dv}{dx} = -\sin x\).

Substituting these into the quotient rule gives:

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

Simplifying the numerator:

\((1 + \cos x)\cos x + \sin^2 x = \cos x + \cos^2 x + \sin^2 x\)

Using the identity \(\cos^2 x + \sin^2 x = 1\), the numerator becomes:

\(\cos x + 1\)

Thus, \(\frac{dy}{dx} = \frac{1}{1 + \cos x}\).

Since \(-1 < \cos x < 1\) for \(-\pi < x < \pi\), \(1 + \cos x > 0\).

Therefore, \(\frac{1}{1 + \cos x} > 0\), showing that the gradient is positive for all \(x\) in the interval.

To find the stationary points, we need to find the derivative of \(y = \frac{{(\ln x)^2}}{x}\) and set it to zero.

Using the quotient rule, if \(u = (\ln x)^2\) and \(v = x\), then \(y = \frac{u}{v}\).

The derivative is given by:

\(\frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}\)

\(\frac{du}{dx} = 2 \ln x \cdot \frac{1}{x} = \frac{2 \ln x}{x}\)

\(\frac{dv}{dx} = 1\)

Substituting these into the quotient rule:

\(\frac{dy}{dx} = \frac{x \cdot \frac{2 \ln x}{x} - (\ln x)^2 \cdot 1}{x^2}\)

\(= \frac{2 \ln x - (\ln x)^2}{x^2}\)

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

\(2 \ln x - (\ln x)^2 = 0\)

\(\ln x (2 - \ln x) = 0\)

This gives \(\ln x = 0\) or \(\ln x = 2\).

If \(\ln x = 0\), then \(x = e^0 = 1\).

If \(\ln x = 2\), then \(x = e^2\).

For \(x = 1\), \(y = \frac{(\ln 1)^2}{1} = 0\).

For \(x = e^2\), \(y = \frac{(\ln e^2)^2}{e^2} = \frac{4}{e^2} = 4e^{-2}\).

Thus, the stationary points are \((1, 0)\) and \(\left(e^2, 4e^{-2}\right)\).

To find the stationary point, we need to differentiate \(y = \\sin x \\cos 2x\) and set the derivative to zero.

Using the product rule, \(\frac{d}{dx}(uv) = u'v + uv'\), where \(u = \\sin x\) and \(v = \\cos 2x\).

Then, \(u' = \\cos x\) and \(v' = -2 \\sin 2x\).

The derivative is:

\(\frac{dy}{dx} = \\cos x \\cos 2x - 2 \\sin x \\sin 2x\).

Using the double angle identity \(\\sin 2x = 2 \\sin x \\cos x\), the derivative becomes:

\(\frac{dy}{dx} = \\cos x \\cos 2x - 4 \\sin x \\cos x \\sin x\).

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

\(\\cos x \\cos 2x - 4 \\sin^2 x \\cos x = 0\).

Factor out \(\\cos x\):

\(\\cos x (\\cos 2x - 4 \\sin^2 x) = 0\).

Since \(\\cos x \neq 0\) in the interval \(0 < x < \frac{1}{2} \pi\), we have:

\(\\cos 2x - 4 \\sin^2 x = 0\).

Using the identity \(\\cos 2x = 1 - 2 \\sin^2 x\), substitute:

\(1 - 2 \\sin^2 x - 4 \\sin^2 x = 0\).

Simplify to get:

\(1 - 6 \\sin^2 x = 0\).

\(6 \\sin^2 x = 1\).

\(\\sin^2 x = \frac{1}{6}\).

\(\\sin x = \frac{1}{\sqrt{6}}\).

Solving for \(x\) in the interval \(0 < x < \frac{1}{2} \pi\), we find:

\(x \approx 0.421\).

To find the stationary point, we need to find where the derivative \(\frac{dy}{dx}\) is zero.

Given \(y = \frac{e^{2x}}{4 + e^{3x}}\), use the quotient rule: \(\frac{dy}{dx} = \frac{(4 + e^{3x})(2e^{2x}) - e^{2x}(3e^{3x})}{(4 + e^{3x})^2}\).

Simplify the numerator: \((4 + e^{3x})(2e^{2x}) - e^{2x}(3e^{3x}) = 8e^{2x} + 2e^{5x} - 3e^{5x} = 8e^{2x} - e^{5x}\).

Set the derivative to zero: \(8e^{2x} - e^{5x} = 0\).

Factor out \(e^{2x}\): \(e^{2x}(8 - e^{3x}) = 0\).

Since \(e^{2x} \neq 0\), solve \(8 - e^{3x} = 0\) to get \(e^{3x} = 8\).

Taking the natural logarithm: \(3x = \ln 8\), so \(x = \frac{\ln 8}{3} = \ln 2\).

Substitute \(x = \ln 2\) back into the original equation to find \(y\):

\(y = \frac{e^{2 \ln 2}}{4 + e^{3 \ln 2}} = \frac{4}{4 + 8} = \frac{4}{12} = \frac{1}{3}\).

Thus, the coordinates of the stationary point are \((\ln 2, \frac{1}{3})\).

To find the stationary points, we need to differentiate \(y = \cos x \cos 2x\) and set the derivative to zero.

Using the product rule, \(\frac{d}{dx}(uv) = u'v + uv'\), where \(u = \cos x\) and \(v = \cos 2x\).

First, find the derivatives: \(u' = -\sin x\) and \(v' = -2\sin 2x\).

Using the product rule: \(\frac{dy}{dx} = (-\sin x)(\cos 2x) + (\cos x)(-2\sin 2x)\).

Simplify: \(\frac{dy}{dx} = -\sin x \cos 2x - 2\cos x \sin 2x\).

Using the double angle formula \(\sin 2x = 2\sin x \cos x\), rewrite \(\cos 2x = 1 - 2\sin^2 x\).

Substitute and simplify: \(\frac{dy}{dx} = -\sin x (1 - 2\sin^2 x) - 4\cos^2 x \sin x\).

Factor out \(\sin x\): \(\frac{dy}{dx} = -\sin x (1 - 2\sin^2 x + 4\cos^2 x)\).

Set \(\frac{dy}{dx} = 0\): \(-\sin x (1 - 2\sin^2 x + 4\cos^2 x) = 0\).

Since \(\sin x = 0\) is not in the interval, solve \(1 - 2\sin^2 x + 4\cos^2 x = 0\).

Using \(\cos^2 x = 1 - \sin^2 x\), substitute: \(1 - 2\sin^2 x + 4(1 - \sin^2 x) = 0\).

Simplify: \(1 - 2\sin^2 x + 4 - 4\sin^2 x = 0\).

Combine terms: \(5 - 6\sin^2 x = 0\).

Solve for \(\sin^2 x\): \(6\sin^2 x = 5\), so \(\sin^2 x = \frac{5}{6}\).

\(\sin x = \sqrt{\frac{5}{6}}\) or \(\sin x = -\sqrt{\frac{5}{6}}\).

Since \(0 < x < \frac{1}{2}\pi\), \(\sin x = \sqrt{\frac{5}{6}}\).

Find \(x\): \(x = \arcsin(\sqrt{\frac{5}{6}}) \approx 1.15\).