Prove a solution of a differential equation is bounded I am interested in the differential equation 
$$
(\mathcal{S})\left\{
\begin{array}{l}
y'(t) = \displaystyle \frac{y^{2}(t)}{1+t^{2}+y^{2}(t)} \\[2mm]
y(0) = \displaystyle \frac{3}{4}
\end{array}
\right.
$$
Let $y$ be a maximal solution of $(\mathcal{S})$. I already proved that $y$ is defined on $\mathbb{R}$. However, I do not see how to prove that $\displaystyle \lim \limits_{t \to +\infty} y(t)$ is lower than $14$ and $\displaystyle \lim \limits_{t \to -\infty} y(t)$ is greater than $\frac{1}{3}$. A hint would be appreciated !

From the equation $0<y'\le1$, so that $y$ is increasing and
$$
\frac34\le y(t)\le\frac34+t,\quad t\ge0.
$$
Substituting into the equation we get
$$
y'\le\frac{y^2}{1+t^2+(3/4)^2}=\frac{y^2}{t^2+(5/4)^2}.
$$
Integrate to obtain
$$
-\frac{1}{y}+\frac43\le\frac45\arctan\frac{4\,t}{5},\quad t\ge0,
$$
and from here
$$
y(t)\le\frac{1}{4/3-(4/5)\arctan t}<\frac{1}{4/3-(4/5)(\pi/2)}=13.038\dots
$$
