If we write the equation as,
$\frac{{dy}}{{dx}} = \frac{{2xy}}{{2{x^2} + {y^2}}}$
and then divide through $x^2$ we will get:
$\frac{{dy}}{{dx}} = \frac{{2\dfrac{y}{x}}}{{2 + {{\left( {\dfrac{y}{x}} \right)}^2}}}$
This suggests that we simplify the previous equation in terms of $f\left( v \right) = \frac{{2v}}{{2 + {v^2}}}$
So putting
\eqalign{ & \frac{y}{x} = v \cr & y = vx \cr & y' = v'x + v \cr}
We get
$\frac{{dv}}{{dx}}x + v = \frac{{2v}}{{2 + {v^2}}}$
Then
$\eqalign{ & \frac{{dv}}{{dx}}x = - \frac{{{v^3}}}{{2 + {v^2}}} \cr & \frac{{dx}}{x} = - \frac{{2 + {v^2}}}{{{v^3}}}dv \cr & \frac{{dx}}{x} = \left( { - \frac{2}{{{v^3}}} - \frac{1}{v}} \right)dv \cr} $
Upon integration we have:
$\log x + C = \frac{1}{{{v^2}}} - \log v$
Let's substitute back
$\eqalign{ & \log x + C = \frac{{{x^2}}}{{{y^2}}} - \log \frac{y}{x} \cr & \log x + C = \frac{{{x^2}}}{{{y^2}}} - \log y + \log x \cr & \log y = \frac{{{x^2}}}{{{y^2}}} - C \cr & y = {C_1}\exp \left( {{x^2}{y^{ - 2}}} \right) \cr} $
You can find $y$ in terms of $x$, but I don't think the inverse is possible, at least with everyday functions.
$y\sqrt {\log y + C} = x$
Ok, using the Lambert W we have
${y^2}\left( {\log y + C} \right) = {x^2}$
Use the exponential:
${e^{{y^2}}}y{e^C} = {e^{{x^2}}}$
Square and multiply by two
$2{y^2}{e^{2{y^2}}}{e^{2C}} = 2{e^{2{x^2}}}$
Use the Lambert W
$2{y^2} = W\left( {\frac{{2{e^{2{x^2}}}}}{{{e^{2C}}}}} \right)$
$y = \sqrt {\frac{1}{2}W\left( {\frac{{2{e^{2{x^2}}}}}{{{e^{2C}}}}} \right)} $
Another aproach would be
$\eqalign{ & \log y + C = \frac{{{x^2}}}{{{y^2}}} \cr & y{e^C} = {e^{\frac{{{x^2}}}{{{y^2}}}}} \cr & {y^2}{e^{2C}} = {e^{2\frac{{{x^2}}}{{{y^2}}}}} \cr & 2\frac{{{x^2}}}{{{y^2}}}{y^2}{e^{2C}} = 2\frac{{{x^2}}}{{{y^2}}}{e^{2\frac{{{x^2}}}{{{y^2}}}}} \cr & 2{x^2}{e^{2C}} = 2\frac{{{x^2}}}{{{y^2}}}{e^{2\frac{{{x^2}}}{{{y^2}}}}} \cr & W\left( {2{x^2}{e^{2C}}} \right) = 2\frac{{{x^2}}}{{{y^2}}} \cr & {y^2} = \frac{{2{x^2}}}{{W\left( {2{x^2}{e^{2C}}} \right)}} \cr & y = \frac{{\sqrt 2 x}}{{\sqrt{W\left( {2{x^2}{e^{2C}}} \right)}}} \cr} $