Show that the solution for $y'=(2-y)(x^2+2y)$ when $y(0)=3$ has the property $y(x) \in (2,3]$ 
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*Without solving the equation, show that the solution for $y'=(2-y)(x^2+2y), y(0)=3$, there exists no $x\in \mathbb{R}$ such that $y(x)=2$.

*Without solving the equation, show that the solution for this problem $y$ has the property that for all $x \in \mathbb{R}$, $2<y(x) \leq 3$


I tried to compare the solution for this problem, $y_0(x)$ with the solution of the same problem with $y(0)=2$, $y_1(x)$, and looked at:
$y_0(x)-y_1(x)=3-2+\int_0^x((2-y_0(t))(t^2+2y_0(t))-(2-y_1(t))(t^2+2y_1(t)))dt=$
$1+\int_0^x(f(t,y_0(t))-f(t,y_1(t)))dt$, when $f(x,y)=(2-y)(x^2+2y)$.
Earlier I was able to prove that in every closed rectangle $D\subseteq \mathbb{R}^2$ $f$ has the Lifschiz property on $D$, and used this to show that $|y_0 (x)-y_1(x)|\leq 1+L\int_0^x|y_0(t)-y_1(t)|dt$ on a rectangle $D$ when $L$ is the Lifschiz constant.
I then falsely assumed that exists some $x_0 \in \mathbb{R}$ such that $y_0(x_0)=2$, in order to say that in a rectangle around $(x_0,2)$ $D$, with an appropriate Lifschiz constant $L$ we have that $|2-y_1(x_0)|\leq 1+L\int_0^{x_0}|y_0(t)-y_1(t)|dt \leq 1$.From here I didn't have a direction.
I also thought of using the fact that if exists $x_0$ such that $y(x_0)=2$ when $y$ solves $y'=(2-y)(x^2+2y), y(0)=3$ then $y'(x_0)=0$, but this didn't yield results either.
I'd like to know what type of direction could be useful for this problem.

Firs of all, we will show that if $y$ is a solution with $y(0)=3$ then there is no $c\in \mathbb{R}$ such that $y(c)=2.$ Assume that there exists. Then the Cauchy problem $y'=(2-y)(x^2+2y), y(c)=2,$ has two different solutions: the solution with $y(0)=3$ and $y(x)=2.$ This contradicts the unicity of solution.
Now, since $y(x)>2, \forall x,$ it follows from $y'=(2-y)(x^2+2y),$ that $y'<0.$ So, $y$ is decreasing. Since $y(0)=3$ we have that $y(x)<3, \forall x\in (0,\infty).$
Note that we cannot say that $y(x)\le 3,\forall x\in\mathbb{R}.$ Since $y(0)=3$ and $y'(x)<0$ we have that $y(x)>3,\forall x\in (-\infty,0).$
