$(1+2e^{x/y})dx+2e^{x/y}(1-x/y)dy=0$
$x=vy\;dx=v\;dy+y\;dv$
$(1+2e^{v})( v\;dy+y\;dv)+2e^{v} (1-v)dy=0$
=> $ (v+2e^v)dy+y(1+2e^v)dv=0$
$\large\frac{dy}{y} +\frac{1+2e^v}{v+2e^v}$$dv=0$
Integrating $\log y +\log (v+2e^v)=\log c$
$x+2ye ^{\large\frac{x}{y} }=c$
Hence a is the correct answer