The ODE is
$xdx+ydy=\frac {a^2 (xdy-ydx)}{x^2+y^2} $
We can take $x^2+y^2$ on the left hand side to get
$(x^2+y^2)d (x^2+y^2)=2a^2 (xdy-ydx) $
I am not able to move ahead from here.
The first thing to do would be to gather all the similar differentials together.
$\displaystyle \big((x^2+y^2)x+2a^2y)\mathrm dx + \big((x^2+y^2)y-2a^2x\big)\mathrm dy)=0$
I would then test to see if it is an exact equation. If it isn't, seek an integrating factor.
I have tried doing this. The problem here is that the DE is not exact and you can only use the integration factor 1/bx+ay where b is the eqn before dx and a is eqn before dy.The integration factor results in an equation with more than 1 term which makes figuring out the differential very unobvious and time consuming.
On the contrary, the equation you get by multiplying both sides by $\displaystyle x^2+ y^2$ is exact.
$\displaystyle (x^2+ y^2)(xdx+ ydy)= a^2(xdy+ ydx)$
$\displaystyle (x^3+ xy^2)dx+ (x^2y+ y^3)dy= a^2xdy+ a^2ydx$
$\displaystyle (x^3+ xy^2- a^2y)dx+ (x^2y+ y^3- a^2x)dy= 0$
The derivative of $\displaystyle x^3+ xy^2- a^2y$ with respect to y is $\displaystyle 2xy- a^2$ and the derivative of $\displaystyle x^2y+ y^3- a^2x$ with respect to x is $\displaystyle 2xy- a^2$. Those are the same so the equation is exact.
$u=x^2 + y^2$
$v = \dfrac x y$
$\dfrac{du}{dx}=2x + 2y \dfrac{dy}{dx}$
$du = 2x dx + 2y dy$
$x dx + y dy = \dfrac 1 2 du$
$\dfrac{dv}{dx} = \dfrac{y - x \frac{dy}{dx}}{y^2}$
$y^2 dv = y dx - x dy$
$\dfrac{u}{1+v^2}dv = ydx - x dy$
$a^2 \dfrac{x dy - y dx}{x^2+y^2} = -a^2 \dfrac{dv}{1+v^2}$
Now we can rewrite the original diff eq as
$\dfrac{du}{2} = -a^2 \dfrac{dv}{1+v^2}$
integrating we get
$\dfrac{u}{2} = -a^2 \arctan(v) + C$
using the original variables and rearranging we get
$\dfrac{x^2 + y^2}{2} +a^2 \arctan\left(\dfrac x y\right) = C$
and this is the implicit solution.