1. Write the differential equation in standard linear form

Given

$$\sin(2x^2)\,\ln(\tan x^2)\,dy + \left(4xy - 4\sqrt2\,x\sin\left(x^2-\frac\pi4\right)\right)dx=0.$$

Rewrite as

$$\sin(2x^2)\ln(\tan x^2)\frac{dy}{dx} + 4xy - 4\sqrt2\,x\sin\left(x^2-\frac\pi4\right)=0.$$

So,

$$\sin(2x^2)\ln(\tan x^2)\frac{dy}{dx}=4\sqrt2\,x\sin\left(x^2-\frac\pi4\right)-4xy.$$

Now use

$$\sin(2x^2)=2\sin(x^2)\cos(x^2),$$

and

$$\sqrt2\sin\left(x^2-\frac\pi4\right)=\sin x^2-\cos x^2.$$

Hence,

$$4\sqrt2\,x\sin\left(x^2-\frac\pi4\right)=4x(\sin x^2-\cos x^2).$$

Also,

$$\sin(2x^2)=2\sin x^2\cos x^2.$$

Therefore,

\frac{dy}{dx}+rac{4x}{\sin(2x^2)\ln(\tan x^2)}y =\frac{4x(\sin x^2-\cos x^2)}{2\sin x^2\cos x^2\ln(\tan x^2)}.

That is,

$$\frac{dy}{dx}+\frac{2x}{\sin x^2\cos x^2\ln(\tan x^2)}y =\frac{2x(\sin x^2-\cos x^2)}{\sin x^2\cos x^2\ln(\tan x^2)}.$$

2. Recognize useful derivatives

Since

$$\tan(x^2)'=2x\sec^2(x^2),$$

we get

$$\frac{d}{dx}\ln(\tan x^2)=\frac{1}{\tan x^2}\cdot 2x\sec^2(x^2) =\frac{2x}{\sin x^2\cos x^2}.$$

Thus the equation becomes

$$\frac{dy}{dx}+\frac{\big(\ln(\tan x^2)\big)'}{\ln(\tan x^2)}y =\frac{\big(\ln(\tan x^2)\big)'}{\ln(\tan x^2)}(\sin x^2-\cos x^2).$$

So it is a linear equation with integrating factor

$$\text{I.F.}=e^{\int \frac{(\ln(\tan x^2))'}{\ln(\tan x^2)}dx} = e^{\ln|\ln(\tan x^2)|} =|\ln(\tan x^2)|.$$

On the given interval and around the points used, we may take the integrating factor as

$$\ln(\tan x^2).$$

3. Multiply throughout by the integrating factor

Then

$$\frac{d}{dx}\left(y\ln(\tan x^2)\right) =\frac{d}{dx}\ln(\tan x^2)\cdot (\sin x^2-\cos x^2).$$

But notice

$$\frac{d}{dx}(\sin x^2-\cos x^2)=2x(\cos x^2+\sin x^2),$$

so a better route is to check for a simpler exact derivative.

Observe from the original rearranged equation:

$$\ln(\tan x^2)\frac{dy}{dx}+\frac{d}{dx}\ln(\tan x^2)\,y =\frac{d}{dx}\ln(\tan x^2)\,(\sin x^2-\cos x^2).$$

Hence

$$\frac{d}{dx}\left(y\ln(\tan x^2)\right) =\frac{d}{dx}\ln(\tan x^2)\,(\sin x^2-\cos x^2).$$

Now simplify the right-hand side:

$$\frac{d}{dx}\ln(\tan x^2)=\frac{2x}{\sin x^2\cos x^2},$$

so

$$\frac{d}{dx}\ln(\tan x^2)(\sin x^2-\cos x^2) =\frac{2x(\sin x^2-\cos x^2)}{\sin x^2\cos x^2}.$$

But

$$\frac{\sin x^2-\cos x^2}{\sin x^2\cos x^2} =\frac{1}{\cos x^2}-\frac{1}{\sin x^2} =\sec x^2-\csc x^2.$$

This suggests direct integration is cumbersome. Instead, try the substitution

$$z=y-(\sin x^2+\cos x^2).$$

Then

$$z'=y'-2x\sin x^2+2x\cos x^2 = y'+2x(\cos x^2-\sin x^2).$$

From the differential equation in linear form,

$$y'+\frac{2x}{\sin x^2\cos x^2\ln(\tan x^2)}y =\frac{2x(\sin x^2-\cos x^2)}{\sin x^2\cos x^2\ln(\tan x^2)}.$$

Now note that

$$\sin x^2\cos x^2(\sec x^2-\csc x^2)=\sin x^2-\cos x^2.$$

So the RHS is exactly

$$\frac{(\ln(\tan x^2))'}{\ln(\tan x^2)}(\sin x^2-\cos x^2).$$

A natural trial solution is

$$y=\sin x^2-\cos x^2.$$

Check it:

$$y'=2x\cos x^2+2x\sin x^2=2x(\sin x^2+\cos x^2).$$

Then

$$y'+\frac{2x}{\sin x^2\cos x^2\ln(\tan x^2)}y =2x(\sin x^2+\cos x^2)+\frac{2x(\sin x^2-\cos x^2)}{\sin x^2\cos x^2\ln(\tan x^2)},$$

which does not match. So instead solve systematically.

4. Use the substitution $t=\ln(\tan x^2)$

Then

$$\frac{dt}{dx}=\frac{2x}{\sin x^2\cos x^2}.$$

The differential equation becomes

$$\frac{dy}{dx}+\frac{t'}{t}y=\frac{t'}{t}(\sin x^2-\cos x^2).$$

Multiply by $t$:

$$t\frac{dy}{dx}+t'y=t'(\sin x^2-\cos x^2).$$

Thus

$$\frac{d}{dx}(ty)=t'(\sin x^2-\cos x^2).$$

Now let $u=x^2$. Then $du=2x\,dx$, and

$$t=\ln(\tan u), \qquad t'\,dx=d(\ln(\tan u)).$$

So

$$ty=\int (\sin u-\cos u)\,d(\ln(\tan u))+C.$$

Use

$$d(\ln(\tan u))=\frac{du}{\sin u\cos u}.$$

Hence

$$\int (\sin u-\cos u)\,d(\ln(\tan u)) =\int \frac{\sin u-\cos u}{\sin u\cos u}\,du =\int (\sec u-\csc u)\,du.$$

Now,

$$\int \sec u\,du=\ln|\sec u+\tan u|,$$ $$\int \csc u\,du=\ln|\csc u-\cot u|.$$

Therefore,

$$\int (\sec u-\csc u)du =\ln|\sec u+\tan u| - \ln|\csc u-\cot u| + C.$$

Using identities,

$$\sec u+\tan u=\frac{1+\sin u}{\cos u}, \qquad \csc u-\cot u=\frac{1-\cos u}{\sin u}.$$

This simplifies, but we do not actually need the full closed form. We use the given points where $u=\frac\pi6$ and $u=\frac\pi3$.

5. Apply the initial condition

At

$$x=\sqrt{\frac\pi6}, \quad u=\frac\pi6,$$

we have

$$y=1,$$

and

$$t=\ln\left(\tan\frac\pi6\right)=\ln\left(\frac1{\sqrt3}\right)=-\frac12\ln 3.$$

So

$$ty=-\frac12\ln 3.$$

Now evaluate the integral from $u=\frac\pi6$ to $u=\frac\pi3$:

$$(ty)\Big|_{\pi/3}-(ty)\Big|_{\pi/6} =\int_{\pi/6}^{\pi/3}(\sec u-\csc u)du.$$

But note the symmetry:

$$\sec\left(\frac\pi2-u\right)=\csc u, \qquad \csc\left(\frac\pi2-u\right)=\sec u.$$

Since the interval $\left[\frac\pi6,\frac\pi3\right]$ is symmetric about $\frac\pi4$, we get

$$\int_{\pi/6}^{\pi/3}\sec u\,du = \int_{\pi/6}^{\pi/3}\csc u\,du.$$

Hence

$$\int_{\pi/6}^{\pi/3}(\sec u-\csc u)du=0.$$

Therefore,

$$(ty)\Big|_{\pi/3}=(ty)\Big|_{\pi/6}=-\frac12\ln 3.$$

At $x=\sqrt{\frac\pi3}$, i.e. $u=\frac\pi3$,

$$t=\ln\left(\tan\frac\pi3\right)=\ln(\sqrt3)=\frac12\ln 3.$$

Thus

$$\frac12\ln 3\cdot y\left(\sqrt{\frac\pi3}\right)=-\frac12\ln 3.$$

So

$$y\left(\sqrt{\frac\pi3}\right)=-1.$$

Hence

$$\left|y\left(\sqrt{\frac\pi3}\right)\right|=1.$$

6. Final answer

$$\boxed{1}$$