$F(x,y) = (x+y)/x$
$F(kx,ky) = k(x+y)/kx = k^0.f(x,y)$
Hence $F(x,y)$ is a homogenous degree of zero.
Step2.
Using the information in the tool box, let us substitute $y = vx and dy/dx = v+xdv/dx.$
$v + x.dv/dx = (x+vx)/x$
Taking x as the common factor and cancelling we get
$v + x.dv/dx = 1 + v$
Bringing v from LHS to the RHS and cancelling wwe get $x.dv/dx = 1$
Step3.
Seperating the variables we get, $dv = dx/x$
integrating on both sides we get, integration of v = integration of dx/x
i.e, $v= logx + C$
substituting for v
$y/x = logx + C$
Therefore, $y = x.logx + Cx$
This is the required solution for the differential equation.