Solution:
\( \sin(2x^{2})\log_{e}(\tan x^{2})\dfrac{dy}{dx}=4\sqrt{2}\,x\sin\!\left(x^{2}-\dfrac{\pi}{4}\right)-4xy \)
\( \therefore \;\dfrac{dy}{dx}+y\left(\dfrac{4x}{\sin(2x^{2})\log_{e}(\tan x^{2})}\right)=\dfrac{4\sqrt{2}\,x\sin\!\left(x^{2}-\dfrac{\pi}{4}\right)}{\sin(2x^{2})\log_{e}(\tan x^{2})} \)
\( \Rightarrow \dfrac{dy}{dx}+y\,P(x)=Q(x) \)
\( I.F.=e^{\displaystyle\int P(x)dx}=e^{\displaystyle\int \dfrac{4x}{\sin(2x^{2})\log_{e}(\tan x^{2})}dx} \)
Let \( \log_{e}(\tan(x^{2}))=t \).
\( \displaystyle\therefore \frac{1}{(\tan x^{2})}\cdot \sec(x^{2})\cdot 2x\,dx=dt \Rightarrow \frac{1}{\sin(x^{2})\log_{e}(x^{2})}(2x)\,dx=dt \)
\(\displaystyle\Rightarrow \frac{4x\,dx}{\sin(2x^{2})}=dt \;\;\therefore\; I.F.=e^{\displaystyle\int \frac{1}{t}dt}=e^{\log_{e}t}=t=\log_{e}(\tan x^{2}) \)
\( \therefore \) Solution, \( y(I.F.)=\displaystyle\int Q(x)\cdot (I.F.)\,dx \)
\( =\displaystyle\int \frac{4x}{\sin(2x^{2})\log_{e}(\tan x^{2})}\times \log_{e}(\tan x^{2})\,dx=\int \frac{4x}{\sin(2x^{2})}dx=\int \frac{d\theta}{\sin\theta}=\text{cosec}\theta\, d\theta \) where \( \theta=2x^{2} \)
\( =\log\!\left|\text{cosec}(2x^{2})-\cot(2x^{2})\right|+c \)
\( \therefore \; y\cdot \log_{e}(\tan x^{2})=\log\!\left|\text{cosec}(2x^{2})-\cot(2x^{2})\right|+c \)
Now as given, \( x=\sqrt{\dfrac{\pi}{6}},\, y=1 \)
\( 1\cdot \log_{e}\!\left(\frac{1}{\sqrt{3}}\right)=\log_{e}\!\left(\frac{2}{\sqrt{3}}-\frac{1}{\sqrt{3}}\right)+c \Rightarrow c=0 \)
So, when \( x=\sqrt{\dfrac{\pi}{3}} \Rightarrow x^{2}=\dfrac{\pi}{3} \Rightarrow 2x^{2}=\dfrac{2\pi}{3} \)
\( y\cdot \log_{e}^\sqrt{3}=\log_{e}\!\left(\dfrac{2}{\sqrt{3}}-\left(-\dfrac{1}{\sqrt{3}}\right)\right)=\log_{e}^\sqrt{3} \)
\( \therefore \; y=1 \)

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