$${{dy} \over {dx}} = {{xy} \over {{x^2} + {y^2}}}$$

Let y = vx

$$ \therefore $$ $${{dy} \over {dx}}$$ = v + x.$${{dv} \over {dx}}$$

$$ \Rightarrow $$ v + x.$${{dv} \over {dx}}$$ = $${{x\left( {vx} \right)} \over {{x^2} + {v^2}{x^2}}}$$ = $${v \over {1 + {v^2}}}$$

$$ \Rightarrow $$ x.$${{dv} \over {dx}}$$ = $${v \over {1 + {v^2}}}$$ - v = $${{v - v - {v^3}} \over {1 + {v^2}}}$$ = $$ - {{{v^3}} \over {1 + {v^2}}}$$

$$ \Rightarrow $$ $$\int {{{1 + {v^2}} \over {{v^3}}}} dv = - \int {{{dx} \over x}} $$

$$ \Rightarrow $$ $$ - {1 \over {2{v^2}}} + \log v$$ = $$ - \log x + C$$

$$ \Rightarrow $$ $$ - {1 \over 2}{{{x^2}} \over {{y^2}}} + \log \left( {{y \over x}} \right)$$ = $$ - \log x + C$$ ......(1)

putting x = 1, y = 1 we get

$$ \Rightarrow $$ C = $$ - {1 \over 2}$$

From eq. (1)

$$ - {1 \over 2}{{{x^2}} \over {{y^2}}} + \log \left( {{y \over x}} \right)$$ = $$ - \log x - {1 \over 2}$$

Put y = e

$$ - {1 \over 2}{{{x^2}} \over {{e^2}}} + \log \left( {{e \over x}} \right)$$ = $$ - \log x - {1 \over 2}$$

$$ \Rightarrow $$ x² = 3e²

$$ \Rightarrow $$ x = $$ \pm $$3$$\sqrt e $$