The function $f$ is defined by $f(x)=3\sin x-4\cos 2x$, $0<x<\dfrac{\pi}{2}$.

1. Find the derivative of $f(x)$

$f'(x) = 3\cos x - 4(-\sin 2x)(2)$
$f'(x) = 3\cos x + 8\sin 2x$

2. Analyse the sign of $f'(x)$ in the given interval

For $0 < x < \frac{\pi}{2}$:
$\cos x > 0$
Since $0 < 2x < \pi$, $\sin 2x > 0$
Therefore, $3\cos x + 8\sin 2x > 0$.

3. Conclude whether the function is increasing or decreasing

Since $f'(x) > 0$ for all $x$ in the interval $0 < x < \frac{\pi}{2}$, $f$ is an increasing function.

1. Find the derivative of $y$ with respect to $x$.

$y = 3\sin x - 4\cos 2x$
$\frac{dy}{dx} = 3\cos x - 4(-\sin 2x)(2)$
$\frac{dy}{dx} = 3\cos x + 8\sin 2x$

2. Evaluate $\frac{dy}{dx}$ at $x = \frac{\pi}{6}$.

When $x = \frac{\pi}{6}$,
$\frac{dy}{dx} = 3\cos\left(\frac{\pi}{6}\right) + 8\sin\left(2 \times \frac{\pi}{6}\right)$
$\frac{dy}{dx} = 3\left(\frac{\sqrt{3}}{2}\right) + 8\sin\left(\frac{\pi}{3}\right)$
$\frac{dy}{dx} = \frac{3\sqrt{3}}{2} + 8\left(\frac{\sqrt{3}}{2}\right)$
$\frac{dy}{dx} = \frac{11\sqrt{3}}{2}$

3. Use the chain rule to find the rate of change of $y$.

Given $\frac{dx}{dt} = 2$,
$\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$
$\frac{dy}{dt} = \frac{11\sqrt{3}}{2} \times 2$
$\frac{dy}{dt} = 11\sqrt{3}$ (shown)