By separation of variables, the general solution to $y' = 3y - y^2$ is
$y(t) = 0$ or $y(t) = \frac{3}{1+ce^{-3t}}$.
Find the interval of validity of the solution satisfying $y(0) = 6$.
(i.e. find $a$ and $b$ such that the solution is valid on the interval $a<t<b$.)
I'm not really sure how to approach this. I set $y' = 0$ and hence $3y - y^2 = 0$, and assuming $y \neq 0$, I proceeded to $3 = y = \frac{3}{1+ce^{-3t}}$. Then $1=1+ce^{-3t}$, which means $c=0$ since any $e^x$ function will never be $0$.
However, this entirely gets rid of the term containing $t$, and doesn't help me find boundaries for $t$.
How should I approach/solve this?
So you found the non-stationary solution of the ODE $y'(t)=3y(t)-y^2(t)$: $$y(t) = \frac{3}{1+ce^{-3t}}$$ where $c$ is a real value different from zero. The interval of validity depends on the initial condition $y(t_0)=y_0$. If $t_0=0$ and $y_0=6$ then $$6=y(0) = \frac{3}{1+ce^{-3\cdot 0}}=\frac{3}{1+c}$$ which implies that $c=-1/2$. Hence the solution that passes through the initial point $(0,6)$ is $$y(t) = \frac{3}{1-\frac{e^{-3t}}{2}}=\frac{6}{2-e^{-3t}}$$ which is not defined for all real values of $t$. The interval of validity of the solution is the biggest interval in the domain which contains $t=0$. Can you take it from here?
P.S. Note that if the initial condition is $y(t_0)=y_0$ with $y_0\in (0,3)$, i.e. between the stationary solutions that it will turn out that the interval of validity is whole real line $\mathbb{R}$. On the other hand, for $y_0$ outside the interval $[0,3]$, as in your case, the interval of validity is just a half-line.