$xdy=(y-\sqrt{x^2 +y^2})dx$
$y(x) =z(x) x$
$dy =zdx+xdz$
$x(zdx+xdz) =(xz-\sqrt{x^2 +x^2 z^2})dx$
...
$xdz =-\sqrt{1^2 +z^2} ×dx$
$\frac {dx} {x} = - \frac {dz}{\sqrt{1^2 +z^2}}$
$ln(x) +C= - ln|z+ \sqrt{1^2 +z^2}|$
...
$x(z+\sqrt{1^2 +z^2}) =C$
$x(\frac{y} {x} + \frac {\sqrt{x^2 +y^2}} {x}) =C$
$y+\sqrt{x^2 +y^2} =C $
Now how to solve it for $y$? And what is $C$ equal to?
$$C-y=\sqrt{x^2+y^2}$$
$$(C-y)^2=x^2+y^2$$
$$y^2-2Cy+C^2=x^2+y^2$$
$$-2Cy+C^2=x^2$$
$$y=\frac{C^2-x^2}{2C}$$
You cannot solve for $C$, as you do not have an initial condition (like in an indefinite integral).