Let $$y = y(x)$$ be a solution of the differential equation $$(x\cos x)dy + (xy\sin x + y\cos x - 1)dx = 0$$, $$0 \lt x \lt \dfrac{\pi}{2}$$. If $$\dfrac{\pi}{3}y\left(\dfrac{\pi}{3}\right) = \sqrt{3}$$, then $$\left|\dfrac{\pi}{6}y''\left(\dfrac{\pi}{6}\right) + 2y'\left(\dfrac{\pi}{6}\right)\right|$$ is equal to ______.

We need to solve the differential equation $$(x\cos x)\,dy + (xy\sin x + y\cos x - 1)\,dx = 0$$ with $$\dfrac{\pi}{3}y\!\left(\dfrac{\pi}{3}\right) = \sqrt{3}$$.

Rewriting in standard linear form gives $$\dfrac{dy}{dx} + y\left(\tan x + \dfrac{1}{x}\right) = \dfrac{\sec x}{x}$$.

The integrating factor is $$\text{IF} = e^{\int (\tan x + 1/x)\,dx} = e^{-\ln \cos x + \ln x} = \dfrac{x}{\cos x} = x\sec x$$.

Multiplying through by $$x\sec x$$, the left-hand side becomes the derivative of $$y \cdot x\sec x$$, so $$\dfrac{d}{dx}(y \cdot x\sec x) = \dfrac{\sec x}{x} \cdot x\sec x = \sec^2 x$$.

Integrating both sides yields $$y \cdot x\sec x = \int \sec^2 x\,dx = \tan x + C$$, which implies $$\dfrac{xy}{\cos x} = \tan x + C \implies xy = \sin x + C\cos x$$.

Applying the initial condition at $$x = \dfrac{\pi}{3}$$ gives $$\dfrac{\pi}{3}\,y\!\left(\dfrac{\pi}{3}\right) = \sin\dfrac{\pi}{3} + C\cos\dfrac{\pi}{3}$$, and since $$\sqrt{3} = \dfrac{\sqrt{3}}{2} + \dfrac{C}{2}$$, we find $$C = \sqrt{3}$$. Thus $$xy = \sin x + \sqrt{3}\cos x$$.

At $$x = \dfrac{\pi}{6}$$, we have $$\dfrac{\pi}{6}\,y\!\left(\dfrac{\pi}{6}\right) = \sin\dfrac{\pi}{6} + \sqrt{3}\cos\dfrac{\pi}{6} = \dfrac{1}{2} + \dfrac{3}{2} = 2$$.

Differentiating $$xy = \sin x + \sqrt{3}\cos x$$ gives $$y + xy' = \cos x - \sqrt{3}\sin x$$, and evaluating at $$x = \dfrac{\pi}{6}$$ yields $$y\!\left(\dfrac{\pi}{6}\right) + \dfrac{\pi}{6}y'\!\left(\dfrac{\pi}{6}\right) = \dfrac{\sqrt{3}}{2} - \dfrac{\sqrt{3}}{2} = 0$$.

Differentiating again gives $$2y' + xy'' = -\sin x - \sqrt{3}\cos x$$, and at $$x = \dfrac{\pi}{6}$$ this becomes $$2y'\!\left(\dfrac{\pi}{6}\right) + \dfrac{\pi}{6}y''\!\left(\dfrac{\pi}{6}\right) = -\dfrac{1}{2} - \dfrac{3}{2} = -2$$.

Therefore, $$\left|\dfrac{\pi}{6}y''\!\left(\dfrac{\pi}{6}\right) + 2y'\!\left(\dfrac{\pi}{6}\right)\right| = |-2| = 2$$, and the answer is $$2$$.