Given,

$$(1 + {e^{2x}}){{dy} \over {dx}} + 2(1 + {y^2}){e^x} = 0$$

$$ \Rightarrow {{dy} \over {dx}} = {{ - 2(1 + {y^2}){e^x}} \over {1 + {e^{2x}}}}$$

$$ \Rightarrow \int {{{dy} \over {1 + {y^2}}} = \int {{{ - 2{e^x}dx} \over {(1 + {e^{2x}})}}} } $$

$$ \Rightarrow {\tan ^{ - 1}}(y) = \int {{{ - 2{e^x}dx} \over {(1 + {e^{2x}})}}} $$

Now,

Let $${e^x} = t$$

$$ \Rightarrow {e^x}dx = dt$$

$$ \Rightarrow {\tan ^{ - 1}}(y) = \int {{{ - 2dt} \over {1 + {t^2}}}} $$

$$ \Rightarrow {\tan ^{ - 1}}(y) = - 2{\tan ^{ - 1}}(t) + C$$

$$ \Rightarrow {\tan ^{ - 1}}(y) = - 2{\tan ^{ - 1}}({e^x}) + C$$ [Putting value of t]

Given, y(0) = 0 means when x = 0 the y = 0

$$\therefore$$ $${\tan ^{ - 1}}(0) = - 2{\tan ^{ - 1}}({e^o}) + C$$

$$ \Rightarrow 0 = - 2 \times {\pi \over 4} + C$$

$$ \Rightarrow C = {\pi \over 2}$$

$$\therefore$$ $${\tan ^{ - 1}}(y) = 2{\tan ^{ - 1}}({e^x}) + {\pi \over 2}$$

$$ \Rightarrow y = \tan \left( { - 2{{\tan }^{ - 1}}({e^x}) + {\pi \over 2}} \right)$$

Differentiating both sides, we get

$$y' = {\sec ^2}\left( { - 2{{\tan }^{ - 1}}({e^x}) + {\pi \over 2}} \right) - 2\,.\,{{{e^x}} \over {1 + {e^{2x}}}}$$

$$ = {\sec ^2}\left( { - 2{{\tan }^{ - 1}}({e^x}) + {\pi \over 2}} \right) \times {{ - 2{e^x}} \over {1 + {e^{2x}}}}$$

$$\therefore$$ $$y'(0) = {\sec ^2}\left( { - 2{{\tan }^{ - 1}}({e^o}) + {\pi \over 2}} \right) \times {{ - 2{e^o}} \over {1 + {e^o}}}$$

$$ = {\sec ^2}\left( { - 2 \times {\pi \over 4} + {\pi \over 2}} \right) \times {{ - 2} \over 2}$$

$$ = {\sec ^2}(0)x - 1$$

$$ = 1 \times 1$$

$$ = - 1$$

And

$$y\left( {\log _e^{\sqrt 3 }} \right) = \tan \left( { - 2{{\tan }^{ - 1}}\left( {{e^{\log _e^{\sqrt 3 }}}} \right) + {\pi \over 2}} \right)$$

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

$$ = \tan \left( { - 2 \times {\pi \over 3} + {\pi \over 2}} \right)$$

$$ = \tan \left( { - {\pi \over 6}} \right)$$

$$ = - \tan \left( {{\pi \over 6}} \right)$$

$$ = - {1 \over {\sqrt 3 }}$$

$$\therefore$$ $$6\left( {y'(0) + {{\left( {y(\log _e^{\sqrt 3 })} \right)}^2}} \right)$$

$$ = 6\left( { - 1 + {{\left( {{{ - 1} \over {\sqrt 3 }}} \right)}^2}} \right)$$

$$ = 6\left( { - 1 + {1 \over 3}} \right) = 6 \times {{ - 2} \over 3} = - 4$$