Solve this : $\,\displaystyle{\frac{dy}{dx}} = \cfrac{2xy \,e^{(x/y)^2}}{y^2(1+e^{(x/y)^2})+2x^2e^{(x/y)^2}}$ 
Solve this : $\,\cfrac{dy}{dx} = \cfrac{2xy \,e^{(x/y)^2}}{y^2(1+e^{(x/y)^2})+2x^2e^{(x/y)^2}}$

I tried to solve it using the homogeneous equation method:
$$y=vx\\ \cfrac{dy}{dx}=v\,+ x\cfrac{dv}{dx}$$
$\implies v\,+x\cfrac{dv}{dx} = \cfrac{2v\,e^{1/v^2}}{v^2(1+e^{1/v^2})+2e^{1/v^2}}$
$\implies x\cfrac{dv}{dx}= \cfrac{-v^2(1+e^{1/v^2})}{v^2(1+e^{1/v^2})+2e^{1/v^2}}$
$\implies \cfrac{v^2(1+e^{1/v^2})+2e^{1/v^2}}{v^2(1+e^{1/v^2})}dv=-\cfrac{dx}{x}$
$\implies \left[1 + \cfrac{2e^{1/v^2}}{v^2(1+e^{1/v^2})}\right]dv = -\cfrac{dx}{x}$
after this I have no clue, please give hint to solve the LHS.
 A: $$\, \cfrac { dy }{ dx } =\cfrac { 2xy\, e^{ (x/y)^{ 2 } } }{ y^{ 2 }(1+e^{ (x/y)^{ 2 } })+2x^{ 2 }e^{ (x/y)^{ 2 } } } \\ \, \cfrac { dy }{ dx } =\cfrac { 2\, e^{ (x/y)^{ 2 } } }{ \frac { y }{ x } (1+e^{ (x/y)^{ 2 } })+2\frac { x }{ y } e^{ (x/y)^{ 2 } } } \\ y=xt\\ \frac { dy }{ dx } =x\frac { dt }{ dx } +t\\ x\frac { dt }{ dx } +t=\frac { 2{ e }^{ \frac { 1 }{ { t }^{ 2 } }  } }{ t\left( 1+{ e }^{ \frac { 1 }{ { t }^{ 2 } }  } \right) +\frac { 2 }{ t } { e }^{ \frac { 1 }{ { t }^{ 2 } }  } } =\frac { 2{ te }^{ \frac { 1 }{ { t }^{ 2 } }  } }{ { t }^{ 2 }\left( 1+{ e }^{ \frac { 1 }{ { t }^{ 2 } }  } \right) +2{ e }^{ \frac { 1 }{ { t }^{ 2 } }  } } \\ x\frac { dt }{ dx } =\frac { 2{ te }^{ \frac { 1 }{ { t }^{ 2 } }  } }{ { t }^{ 2 }\left( 1+{ e }^{ \frac { 1 }{ { t }^{ 2 } }  } \right) +2{ e }^{ \frac { 1 }{ { t }^{ 2 } }  } } -t=\frac { -{ t }^{ 3 }-{ t }^{ 3 }{ e }^{ \frac { 1 }{ { t }^{ 2 } }  } }{ { t }^{ 2 }\left( 1+{ e }^{ \frac { 1 }{ { t }^{ 2 } }  } \right) +2{ e }^{ \frac { 1 }{ { t }^{ 2 } }  } } \\ \int { \frac { { t }^{ 2 }\left( 1+{ e }^{ \frac { 1 }{ { t }^{ 2 } }  } \right) +2{ e }^{ \frac { 1 }{ { t }^{ 2 } }  } }{ { t }^{ 3 }\left( 1+{ e }^{ \frac { 1 }{ { t }^{ 2 } }  } \right)  } dt } =-\int { \frac { dx }{ x }  } \\ \int { \frac { dt }{ t } -\int { \frac { d\left( { 1+e }^{ \frac { 1 }{ { t }^{ 2 } }  } \right)  }{ \left( 1+{ e }^{ \frac { 1 }{ { t }^{ 2 } }  } \right)  }  }  } =-\ln { \left| x \right| +C } \\ \ln { \left| t \right| -\ln { \left| 1+{ e }^{ \frac { 1 }{ { t }^{ 2 } }  } \right| = }  } -\ln { \left| x \right| +C } \\ $$ 

$$\ln { \left| \frac { y }{ \left( 1+e^{ (x/y)^{ 2 } } \right)  }  \right|  } =C\\ y=C\left( 1+e^{ (x/y)^{ 2 } } \right) \\ \\ \\ \\ $$

A: Subs $\ x=uy \ $ in your diff. equation:
$\,\displaystyle{\frac{dy}{dx}} = \cfrac{2xy \,e^{(x/y)^2}}{y^2(1+e^{(x/y)^2})+2x^2e^{(x/y)^2}}$
then $dx=ydu+udy$ and
$$ \,\displaystyle{\frac{dy}{dx}} = \cfrac{2uy^{2} \,e^{u^2}}{y^2(1+e^{u^2})+2u^2y^{2}e^{u^2}} $$
Inverse the fraction in both sides to get:
$$ \displaystyle{\frac{dx}{dy}} = \frac{y^2(1+e^{u^2})+2u^2y^{2}e^{u^2}}{2uy^{2} \,e^{u^2}} $$
this implies that
$$ \,y\frac{du}{dy}+u=\frac{ydu+udy}{dy}=\displaystyle{\frac{dx}{dy}} = \frac{y^2(1+e^{u^2})+2u^2y^{2}e^{u^2}}{2uy^{2} \,e^{u^2}} = \frac{(1+e^{u^2})+2u^2e^{u^2}}{2u \,e^{u^2}}$$ 
and then you can try to solve $u$ as a function of $y$. You can solve the last equation as a separable equation in $u$ and $y$. Put the $y$'s in one side and the $u$'s in the other side and do integration.
A: $$\,\cfrac{dy}{dx} = \cfrac{2xy \,e^{(x/y)^2}}{y^2(1+e^{(x/y)^2})+2x^2e^{(x/y)^2}}\implies
\cfrac{dx}{dy} = \cfrac{y^2(1+e^{(x/y)^2})+2x^2e^{(x/y)^2}}{2xy \,e^{(x/y)^2}}$$
To get rid of the exponential term, let
$$x=y\sqrt{\log( u)}\implies \cfrac{dx}{dy}=\sqrt{\log( u)}+\frac{y \cfrac{du}{dy}} {2u \sqrt{\log( u)}}$$ which makes
$$\sqrt{\log( u)}+\frac{y \cfrac{du}{dy}} {2u \sqrt{\log( u)}}=\sqrt{\log (u)}+\frac{1}{2 u \sqrt{\log (u)}}+\frac{1}{2 \sqrt{\log (u)}}$$ which reduces to 
$$y \cfrac{du}{dy}=1+u\implies \frac{du}{1+u}=\frac{dy}y$$ which is quite simple.
