Official solution
Given differential equation
\[2y{e^{{x \over {{y^2}}}}}dx + \left( {{y^2} - 4x{e^{{x \over {{y^2}}}}}} \right)dy = 0,\,x(1) = 0\]
\[ \Rightarrow {e^{{x \over {{y^2}}}}}[2ydx - 4xdy] = - {y^2}dy\]
\[ \Rightarrow {e^{{x \over {{y^2}}}}}\left[ {{{2{y^2}dx - 4xydy} \over {{y^4}}}} \right] = {{ - 1} \over y}dy\]
\[ \Rightarrow 2{e^{{x \over {{y^2}}}}}d\left( {{x \over {{y^2}}}} \right) = - {1 \over y}dy\]
\[ \Rightarrow 2{e^{{x \over {{y^2}}}}} = - \ln y + c\] ...... (i)
Now, using x(1) = 0, c = 2
So, for x(e), Put y = e in (i)
\[2{e^{{x \over {{e^2}}}}} = - 1 + 2\]
\[ \Rightarrow {x \over {{e^2}}} = \ln \left( {{1 \over 2}} \right) \Rightarrow x(e) = - {e^2}\ln 2\]