Let $(x,y)$ be the maximal solution of $\begin{cases}x' = cos(x) (t-y^2) \\ y' = xy\end{cases}$ with $x(0) = 0$ and $y(0) = 1$.
I'm asked to prove that $|x(t)| < \frac \pi 2, 0 < y(t) < e^{\frac \pi 2 t}$ in the domain of $(x,y)$.
Let $(x,y)$ be the maximal solution of $\begin{cases}x' = cos(x) (t-y^2) \\ y' = xy\end{cases}$ with $x(0) = 0$ and $y(0) = 1$.
I'm asked to prove that $|x(t)| < \frac \pi 2, 0 < y(t) < e^{\frac \pi 2 t}$ in the domain of $(x,y)$.
Any solution that starts in one of the planes $\{(t,x,y):\cos x=0\}$ stays inside that plane.
No other solution can can cross these planes by the uniqueness theorem.
Initially, $x(0)=0\in(-\frac\pi2,\frac\pi2)$, so that $x(t)\in(-\frac\pi2,\frac\pi2)$ for the full solution.
The bound on $y$ follows, you might cite the Grönwall lemma for this.
The surface $\{(t,x,y):y^2-t\}=0$ has no influence on the claims, but the sign of the factor will influence the long-term behavior of the solution inside these bounds.