Rewrite: $y' = 1 + y/x$. This is homogeneous. Let $v = y/x$, so $y = vx$, $y' = v + xv'$. Then $v + xv' = 1+v \Rightarrow xv' = 1 \Rightarrow v' = 1/x \Rightarrow v = \ln|x| + C$. Therefore $y = vx = x(\ln|x|+C) = x\ln|x| + Cx = x\ln|kx|$ where $k = e^C$.