Doesn't you textbook define "homogeneous" for first order differential equations? (Note that the use of "homogenous" for first order equation and higher equations are quite different.)

Any first order differential equation can be written M(x, y)dx+ N(x, y)dy= 0 and is said to be "homogeneous" if M and N are homogeneous (all terms are powers of x and y that add to the same power like "3x^4+ x^2y^2+ xy^3") of the same degree, for example \(\displaystyle x^2dx+ xydy= 0\) is "homogeneous of degree 2"?

But higher order, linear, differential equations are "homogenous" if every term has y or derivative of y term. For example \(\displaystyle y''- 3y'+ 4y= 0\) is "homogeneous" but \(\displaystyle y''- 3y'+ 4y= 3x\) is not.

If what you are really asking is for solutions of y'+ 3y= 1 then, yes, you can write it as \(\displaystyle \frac{dy}{dx}= 1- 3y\) and then \(\displaystyle \frac{dy}{1- 3y}= dx\) and integrating both sides, \(\displaystyle -\frac{1}{3}\ln(1- 3y)= x+ c\).

You dropped the "dx" from the right side of the equation!

(we could divide both sides by -3 to get \(\displaystyle \ln(1- 3y)= -3x+ c\), \(\displaystyle 1- 3y= e^{-3x}e^{c}= Ce^{-3x}\) where \(\displaystyle C= e^c\). Then \(\displaystyle -3y= -1+ Ce^{-3x}\) and \(\displaystyle y= \frac{1}{3}+ c'e^{-3x}\) where \(\displaystyle c'= -C/3\).)

Another way to solve this equation is to first solve the equation y'+ 3y= 0("homogeneous in the sense usually used for higher order equations) which can be written as \(\displaystyle \frac{dy}{dx}= -3y\) and \(\displaystyle \frac{dy}{y}= -3dx/\). Integrating, \(\displaystyle ln(y)= -3x+ C\) so \(\displaystyle y= C'e^{-3x}\). Now we look for a constant solution, y= c, to the entire equation, y'+ 3y= 1 so that y'= 0 and the equation becomes 3c= 1 so c= 1/3 and the entire solution is \(\displaystyle y= C'e^{-3x}+ \frac{1}{3}\) just as before.

As I said before, your mistake is that you completely dropped the "dx"! The solution you say was given by "online solvers", -ln(|3y - 1|) / 3 + c, is also wrong because it is NOT equal to anything! Was the answer possibly x= -ln(|3y - 1|) / 3 + c?