Characterization of solutions to $f' = f(1-f)$

writing your equation in the form $$-\frac{\frac{df(x)}{dx}}{(f(x)-1)(f(x)}=1$$ and by integrating $$-\int \frac{\frac{df(x)}{dx}}{(f(x)-1)f(x)}dx=\int 1dx$$ doing this we obtain $$-\log(-f(x)+1)+\log(f(x))=x+C$$

Note that since $f(0)=1/2$ then, $f\not\equiv 1$ and , $f\not\equiv 0$ hence a solution cannot an equilibrium $$f'=f(1-f)\Longleftrightarrow\frac{f'}{f} +\frac{f'}{1-f} =1\Longleftrightarrow \ln (|f|)- \ln(|1-f|) =x+c$$

But, $f(x)\in (0,1)$ we obtain $$ \ln (f)- \ln(1-f) =x+c\Longleftrightarrow \frac{f}{1-f} =ke^x \Longleftrightarrow f(x) =1-\frac{1}{1+ ke^x}$$

I am sure you can the $k$ by yourself.