$$\text{Let } y = y(x) \text { be the solution of the differential equation } x\frac{dy}{dx} - y = x^2 \cot x, \quad x \in (0, \pi).\text{ If } y\left(\frac{\pi}{2}\right) = \frac{\pi}{2}, \text{ then } 6y\left(\frac{\pi}{6}\right) - 8y\left(\frac{\pi}{4}\right) \text{ is equal to :}$$

The given differential equation is:

$$x\frac{dy}{dx} - y = x^2 \cot x, \quad x \in (0, \pi)$$

with initial condition $$y\left(\frac{\pi}{2}\right) = \frac{\pi}{2}$$.

First, rewrite the equation in standard linear form:

$$\frac{dy}{dx} - \frac{1}{x}y = x \cot x$$

Here, $$P(x) = -\frac{1}{x}$$ and $$Q(x) = x \cot x$$.

The integrating factor (I.F.) is:

$$\text{I.F.} = e^{\int P(x) dx} = e^{\int -\frac{1}{x} dx} = e^{-\ln x} = \frac{1}{x} \quad (\text{since } x > 0)$$

Multiply both sides by I.F.:

$$\frac{1}{x} \frac{dy}{dx} - \frac{1}{x^2} y = \cot x$$

The left side is the derivative of $$\frac{y}{x}$$:

$$\frac{d}{dx}\left( \frac{y}{x} \right) = \cot x$$

Integrate both sides:

$$\int \frac{d}{dx}\left( \frac{y}{x} \right) dx = \int \cot x dx$$

$$\frac{y}{x} = \ln |\sin x| + C$$

Since $$x \in (0, \pi)$$, $$\sin x > 0$$, so:

$$\frac{y}{x} = \ln (\sin x) + C$$

Thus, the general solution is:

$$y = x \ln (\sin x) + C x$$

Apply the initial condition $$y\left(\frac{\pi}{2}\right) = \frac{\pi}{2}$$:

Substitute $$x = \frac{\pi}{2}$$, $$y = \frac{\pi}{2}$$:

$$\frac{\pi}{2} = \frac{\pi}{2} \ln \left( \sin \frac{\pi}{2} \right) + C \cdot \frac{\pi}{2}$$

$$\sin \frac{\pi}{2} = 1$$, $$\ln(1) = 0$$, so:

$$\frac{\pi}{2} = 0 + \frac{C\pi}{2}$$

$$C = 1$$

Therefore, the particular solution is:

$$y = x \ln (\sin x) + x = x \left( \ln (\sin x) + 1 \right)$$

Now compute $$y\left(\frac{\pi}{6}\right)$$ and $$y\left(\frac{\pi}{4}\right)$$:

For $$x = \frac{\pi}{6}$$:

$$\sin \frac{\pi}{6} = \frac{1}{2}$$

$$y\left(\frac{\pi}{6}\right) = \frac{\pi}{6} \left( \ln \left( \frac{1}{2} \right) + 1 \right) = \frac{\pi}{6} \left( -\ln 2 + 1 \right)$$

For $$x = \frac{\pi}{4}$$:

$$\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$$

$$y\left(\frac{\pi}{4}\right) = \frac{\pi}{4} \left( \ln \left( \frac{1}{\sqrt{2}} \right) + 1 \right) = \frac{\pi}{4} \left( \ln (2^{-1/2}) + 1 \right) = \frac{\pi}{4} \left( -\frac{1}{2} \ln 2 + 1 \right)$$

Now evaluate the expression:

$$6y\left(\frac{\pi}{6}\right) - 8y\left(\frac{\pi}{4}\right) = 6 \left[ \frac{\pi}{6} (1 - \ln 2) \right] - 8 \left[ \frac{\pi}{4} \left(1 - \frac{1}{2} \ln 2\right) \right]$$

Simplify:

$$= \pi (1 - \ln 2) - 2\pi \left(1 - \frac{1}{2} \ln 2\right)$$

Factor out $$\pi$$:

$$= \pi \left[ (1 - \ln 2) - 2 \left(1 - \frac{1}{2} \ln 2\right) \right]$$

Distribute the $$-2$$:

$$= \pi \left[ 1 - \ln 2 - 2 + 2 \cdot \frac{1}{2} \ln 2 \right] = \pi \left[ 1 - \ln 2 - 2 + \ln 2 \right]$$

Combine like terms:

$$= \pi \left[ (1 - 2) + (-\ln 2 + \ln 2) \right] = \pi \left[ -1 + 0 \right] = -\pi$$

Thus, $$6y\left(\frac{\pi}{6}\right) - 8y\left(\frac{\pi}{4}\right) = -\pi$$.

The correct answer is option D: $$-\pi$$.