$${{dy} \over {dx}} + y\left( {{{4x} \over {\sin (2{x^2})\ln (\tan {x^2})}}} \right) = {{4\sqrt 2 x\sin \left( {{x^2} - {\pi \over 4}} \right)} \over {\sin (2{x^2})\ln (\tan {x^2})}}$$

$$I.F = {e^{\int {{{4x} \over {\sin (2{x^2})\ln (\tan {x^2})}}dx} }}$$

$$ = {e^{\ln |\ln (\tan {x^2})}} = \ln (\tan {x^2})$$

$$\therefore$$ $$\int {d(y.\ln (\tan {x^2})) = \int {{{4\sqrt 2 x\sin \left( {{x^2} - {\pi \over 4}} \right)} \over {\sin (2{x^2})}}dx} } $$

$$ \Rightarrow y\ln (\tan {x^2}) = \ln \left| {{{\sec {x^2} + \tan {x^2}} \over {{\mathop{\rm cosec}\nolimits} \,{x^2} - \cot {x^2}}}} \right| + C$$

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

$$e = \ln \left( {{1 \over {\sqrt 3 }}} \right) - \ln \left( {{{\sqrt 3 } \over {2 - \sqrt 3 }}} \right)$$

For $$y\left( {\sqrt {{\pi \over 3}} } \right)$$

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

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

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

$$ \Rightarrow {y \over 2}\ln 3 = - {1 \over 2}\ln 3$$

$$ \Rightarrow y = 1$$

$$\therefore$$ $$\left| {y\left( {\sqrt {{\pi \over 3}} } \right)} \right| = 1$$.