Why is it true that the solution to this IVP is always $<3$ when $x\geq 0$ (without solving the equation)? I'm doing some more practice problems for my upcoming DEs test, and I tried this true/false question:

The solution to the initial value problem $dy/dx=(x-2)(y-3)^2,y(0)=0$, will always be less than $3$; that is, $y(x)<3$ for $x\geq 0$.

Bear in mind that the question assumes the person doing the problem doesn't know how to solve DEs yet, since that is covered in the subsequent chapter. Using the uniqueness and exactness theorem, given that $dy/dx=F(x,y)$, I checked if $F$ and $\partial F/\partial y$ were continuous at the point $(0,0)$ by inputting the point $(0,0)$ into both. If they are continuous, there exists a rectangle $R$ that includes the point $(0,0)$ in which $F$ and $\partial F/\partial y$ are continuous.
$F(0,0)=-2(-3)^2=-18$
$\displaystyle\frac{\partial F}{\partial y}=2(x-2)(y-3)$
$\displaystyle\frac{\partial F}{\partial y}\bigg|_{(0,0)}=2(-2)(-3)=12$
Since the limits of $F$ and $\partial F/\partial y$ can be solved via substitution, the limits are the same as the evaluations of the points, and they are both continuous at any point. Therefore, there exists a rectangle $R$ that includes $(0,0)$ where both are continuous, meaning there is a unique solution to the initial value problem. However, I at first concluded that since $\partial F/\partial y$ is positive, $y$ is increasing at $(0,0)$, and the statement is false. However, the answer is that the statement is true. I've thought about it a bit more and realized that since $dy/dx$ is negative at that point, $y$ is actually decreasing at $(0,0)$, not increasing! Also, we can't conclude anything about the statement from the values at $(0,0)$ besides the fact that only one unique solution exists. So I tried setting $dy/dx$ to $0$ to see if I could find any critical points, in an attempt to see if there's a maximum at $y=3$.
$dy/dx=(x-2)(y-3)^2=0 \implies x=2\;\text{or}\;y=3\implies$ Critical value at $y=3$.
But the second derivative test was inconclusive:
$\begin{align}\displaystyle\frac{d^2 y}{dx^2}&=(y-3)^2+2(x-2)(y-3)\displaystyle\frac{dy}{dx}\\
&=(y-3)^2+2(x-2)(y-3)(x-2)(y-3)^2\\
&=(y-3)^2+2(x-2)^2(y-3)^3\\
\displaystyle\frac{d^2y}{dx^2}\bigg|_{y=3}&=(3-3)^2+(x-2)^2(3-3)^3\\
&=0
\end{align}$
So that didn't work!
However, I just tried solving the DE to see what would happen since the test I'm taking covers solving separable DEs anyway:
$\begin{align}
\displaystyle\frac{dy}{dx}&=(x-2)(y-3)^2 \\
\displaystyle\int\frac{dy}{(y-3)^2}&=\int(x-2)dx\\
-\displaystyle\frac{1}{y-3}&=x^2-2x+C \\
\end{align}$
Applying the initial condition:
$-\displaystyle\frac{1}{0-3}=(0)^2-2(0)+C\implies C=\displaystyle\frac{1}{3}$.
Therefore the unique solution is $y=-\displaystyle\frac{1}{x^2-2x+\frac{1}{3}}+3$
It can now be easily seen that the maximum of this expression is $y=3$ since $\lim_{x\to \infty} y(x)=3$. But is there any way to determine this without solving the equation?
 A: Assume that there exists $x_0>0$ such that $y(x_0)>3$. Then, as $y$ is continuous and $y(0)=0$, we get by the intermediate value theorem that there exists $x_1>0$ such that $y(x_1)=3$. Now we see that $y$ solves the IVP
$$ \begin{cases} \frac{dy}{dx} &= (x-2)(y-3)^2;\\ y(x_1)&=3. \end{cases} $$
However, also $u\equiv 3$ solves the IVP above. Hence, by local uniqueness we know that there exists $\varepsilon>0$ such that $y(x) = 3$ for $x\in (x_1-\varepsilon, x_1+\varepsilon)$.
Now we want to show that in fact $y(x)=3$ for all $x\geq 0$. For this we define
$$ S=\{ x\in \mathbb{R} \ : \ 0\leq x \leq x_1, y(x) = 3 \}. $$
We want to show that $S=[0;x_1]$. For this it is enough to show that $S$ is both open, closed and nonempty. Nonempty is clear as $x_1\in S$. Closed follows from continuity of $y$ and openness follows from redoing the argument above with the local uniqueness theorem.
Hence, we have shown that once $y$ attains $3$, it has always been equal to $3$ (and in fact it would also always be $3$ in the future). Thus, this yields a contradiction to the initial condition $y(0)=0$.
Note that we have not used the particular form of the ODE. If you are given any ODE $\frac{dy}{dx}=F(x,y)$ with $F$ locally lipschitz with $F(\cdot ,y_0)=0$, then the very same argument as above tell us that if $y$ attains $y_0$, then the solution must be constant for all times.
