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
$$\eqalign{ & \log y + C = \frac{{{x^2}}}{{{y^2}}} \cr & y{e^C} = {e^{\frac{{{x^2}}}{{{y^2}}}}} \cr & {y^2}{e^{2C}} = {e^{2\frac{{{x^2}}}{{{y^2}}}}} \cr & 2\frac{{{x^2}}}{{{y^2}}}{y^2}{e^{2C}} = 2\frac{{{x^2}}}{{{y^2}}}{e^{2\frac{{{x^2}}}{{{y^2}}}}} \cr & 2{x^2}{e^{2C}} = 2\frac{{{x^2}}}{{{y^2}}}{e^{2\frac{{{x^2}}}{{{y^2}}}}} \cr & W\left( {2{x^2}{e^{2C}}} \right) = 2\frac{{{x^2}}}{{{y^2}}} \cr & {y^2} = \frac{{2{x^2}}}{{W\left( {2{x^2}{e^{2C}}} \right)}} \cr & y = \frac{{\sqrt 2 x}}{{\sqrt{W\left( {2{x^2}{e^{2C}}} \right)}}} \cr} $$