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

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

$\Rightarrow \ln \left(\tan x^2\right)dy+\frac{4xydx}{\sin \left(2x^2\right)}-\frac{4\sqrt{2}x\sin \left(x^2-\frac{\pi }{4}\right)}{\sin \left(2x^2\right)}dx=0$

$\Rightarrow d\left(y·\ln \left(\tan x^2\right)\right)-4\sqrt{2}x\frac{\left(\sin x^2-\cos x^2\right)}{\sqrt{2}\left(2\sin x^2\cos x^2\right)}dx=0$

$\Rightarrow d\left(y\ln \left(\tan x^2\right)\right)-\frac{4x\left(\sin x^2-\cos x^2\right)}{{\left(\sin x^2+{\cos }^{2}x\right)}^2-1}dx=0$

Now integrating both side,

$\Rightarrow \int d\left(y\ln \left(\tan x^2\right)\right)-\int \frac{4x\left(\sin x^2-\cos x^2\right)}{{\left(\sin x^2+{\cos }^{2}x\right)}^2-1}dx=\int 0$

Now let $\sin x^2+{\cos }^{2}x=t\Rightarrow -2x\left(\sin x^2-{\cos }^{2}x\right)=dt$

$\Rightarrow \int d\left(y\ln \left(\tan x^2\right)\right)+2\int \frac{dt}{t^2-1}=\int 0$

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

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

Put $y=1$ and $x=\sqrt{\frac{\pi }{6}}$

$1\ln \left(\frac{1}{\sqrt{3}}\right)+\ln \frac{\left(\frac{1}{2}+\frac{\sqrt{3}}{2}-1\right)}{\left(\frac{1}{2}+\frac{\sqrt{3}}{2}+1\right)}=c$

Now at $x=\sqrt{\frac{\pi }{3}}$

$\Rightarrow y\left(\ln \sqrt{3}\right)+\ln \frac{\left(\frac{\sqrt{3}}{2}+\frac{1}{2}-1\right)}{\left(\frac{\sqrt{3}}{2}+\frac{1}{2}+1\right)}=\ln \left(\frac{1}{\sqrt{3}}\right)+\ln \left(\frac{\sqrt{3}-1}{\sqrt{3}+3}\right)$

$\Rightarrow y\left(\ln \sqrt{3}\right)=\ln \left(\frac{1}{\sqrt{3}}\right)$

$\Rightarrow y=-1$

So, $\left|y\right|=1$