if you multiply by dx you get:

xdy = ydx + sqrt(x^2 +y^2)dx

(y + sqrt(x^2+y^2))dx -xdy = 0

y =vx

using dy = vdx + xdv => (vx + sqrt(x^2+v^2x^2)dx - x(vdx+xdv) = 0

(vx + x sqrt(1+v^2) -xv)dx - x^2dv = 0

(xsqrt(1+v^2))dx -x^2 dv = 0

which now separates 1/x dx = dv/sqrt(1+v^2)

Typically this is not how you would approach this problem--no need to put

in the form Mdx + Ndy = 0.

dy/dx = y/x + sqrt[1+(y/x)^2]

simply y = vx dy/dx = v + xdv/dx

v+ xdv/dx = v +sqrt(1+v^2)

dv/sqrt(1+v^2) = dx/x

By the way if dy/dx = f(y/x) we call it a differnetial equation with homogeneous coefficients which can be solved by letting y = vx