Let $$y=y(x)$$ be the solution curve of the differential equation $$\sin(2x^2)\log_e(\tan x^2)\,dy+\left(4xy-4\sqrt2\,x\sin x^2-\frac{\pi}{4}\right)dx=0,$$ $$0<x<\sqrt{\frac{\pi}{2}},$$ which passes through the point $$\left(\sqrt{\frac{\pi}{6}},1\right).$$ Then $$y\left(\sqrt{\frac{\pi}{3}}\right)$$ is equal to ______.

Given differential equation,

$$\sin2x^2\ln(\tan x^2)\,dy+\left(4xy-4\sqrt2\,x\sin x^2-\frac{\pi}{4}\right)dx=0$$

Rearranging,

$$\ln(\tan x^2)\,dy+\frac{4xy-4\sqrt2\,x\sin x^2-\frac{\pi}{4}}{\sin2x^2}\,dx=0$$

Using

$$\sin2x^2=2\sin x^2\cos x^2,$$

$$dy\cdot\ln(\tan x^2)+\frac{4xy}{\sin2x^2}dx-\frac{4\sqrt2\,x\sin x^2-\frac{\pi}{4}}{\sin2x^2}dx=0$$

Now,

$$\frac{4x}{\sin2x^2}=\frac{2x}{\sin x^2\cos x^2}$$

Hence,

$$dy\cdot\ln(\tan x^2)-4\sqrt2\,x\frac{\sin x^2-\cos x^2}{\sin^2x^2+\cos^2x^2-1}\,dx=0$$

Therefore,

$$\int dy\,\ln(\tan x^2)+2\int\frac{dt}{t^2-1}=0$$

Integrating,

$$y\ln(\tan x^2)+2\cdot\frac12\ln\left(\frac{t-1}{t+1}\right)=c$$

$$y\ln(\tan x^2)+\ln\left(\frac{\sin x^2+\cos x^2-1}{\sin x^2+\cos x^2+1}\right)=c$$

Using the point

$$\left(\sqrt{\frac{\pi}{6}},1\right),$$

we get

$$\ln\frac1{\sqrt3}+\ln\left(\frac{\frac12+\frac{\sqrt3}{2}-1}{\frac12+\frac{\sqrt3}{2}+1}\right)=c$$

Now put

$$x=\sqrt{\frac{\pi}{3}}$$

Then

$$\tan x^2=\tan\frac{\pi}{3}=\sqrt3$$

Hence,

$$y(\ln\sqrt3)+\ln\left(\frac{\frac12+\frac{\sqrt3}{2}-1}{\frac12+\frac{\sqrt3}{2}+1}\right)=\ln\frac1{\sqrt3}+\ln\left(\frac{\sqrt3-1}{\sqrt3+3}\right)$$

Therefore,

$$y(\ln\sqrt3)=\ln\frac1{\sqrt3}$$

$$y=-1$$

Hence,

$$|y|=1$$

Therefore, the required answer is

$$\boxed{1}$$.