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

$\begin{array}{l}\sin \left(2{\mathrm{x}}^2\right)\ln \left(\tan {\mathrm{x}}^2\right)\mathrm{dy}+\left(4\mathrm{xy}-4\sqrt{2}\mathrm{xsin}\left({\mathrm{x}}^2-\frac{\mathrm{\pi }}{4}\right)\right)\mathrm{dx}=0\\ \ln \left(\tan {\mathrm{x}}^2\right)\mathrm{dy}+\frac{4\mathrm{xydx}}{\sin \left(2{\mathrm{x}}^2\right)}-\frac{4\sqrt{2}\mathrm{xsin}\left({\mathrm{x}}^2-\frac{\mathrm{\pi }}{4}\right)}{\sin \left(2{\mathrm{x}}^2\right)}\mathrm{dx}=0\\ \mathrm{d}\left(\mathrm{y}\cdot \ln \left(\tan {\mathrm{x}}^2\right)\right)-4\sqrt{2}\mathrm{x}\frac{\left(\sin {\mathrm{x}}^2-\cos {\mathrm{x}}^2\right)}{\sqrt{2}\left(2\sin {\mathrm{x}}^2\cos {\mathrm{x}}^2\right)}\mathrm{dx}=0\\ \mathrm{d}\left(\mathrm{yln}\left(\tan {\mathrm{x}}^2\right)\right)-\frac{4\mathrm{x}\left(\sin {\mathrm{x}}^2-\cos {\mathrm{x}}^2\right)}{{\left(\sin {\mathrm{x}}^2+\cos x^2\right)}^2-1}\mathrm{dx}=0\end{array}$

$\begin{array}{l}\Rightarrow \int \mathrm{d}\left(\mathrm{yln}\left(\tan {\mathrm{x}}^2\right)\right)+2\int \frac{\mathrm{dt}}{{\mathrm{t}}^2-1}=\int 0\\ \Rightarrow \mathrm{yln}\left(\tan {\mathrm{x}}^2\right)+2\cdot \frac{1}{2}\ln \left|\frac{\mathrm{t}-1}{\mathrm{t}+1}\right|=\mathrm{c}\\ \mathrm{yln}\left(\tan {\mathrm{x}}^2\right)+\ln \left(\frac{\sin {\mathrm{x}}^2+\cos {\mathrm{x}}^2-1}{\sin {\mathrm{x}}^2+\cos {\mathrm{x}}^2+1}\right)=\mathrm{c}\\ \text{ Put }\mathrm{y}=1\text{ and }\mathrm{x}=\sqrt{\frac{\mathrm{\pi }}{6}}\\ 1\ln \left(\frac{1}{\sqrt{3}}\right)+\ln \frac{\left(\frac{1}{2}+\frac{\sqrt{3}}{2}-1\right)}{\left(\frac{1}{2}+\frac{\sqrt{3}}{2}+1\right)}=\mathrm{c}\\ \text{ Now }\mathrm{x}=\sqrt{\frac{\mathrm{\pi }}{3}}\Rightarrow \mathrm{y}(\ln \sqrt{3})+\ln \frac{\left(\frac{1}{2}+\frac{\sqrt{3}}{2}-1\right)}{\left(\frac{1}{2}+\frac{\sqrt{3}}{2}+1\right)}=\ln \left(\frac{1}{\sqrt{3}}\right)+\ln \left(\frac{\sqrt{3}-1}{\sqrt{3}+3}\right)\\ \mathrm{y}(\ln \sqrt{3})=\ln \left(\frac{1}{\sqrt{3}}\right)\\ \Rightarrow \mathrm{y}=-1\\ \left|\mathrm{y}\right|=1\end{array}$