Consider \begin{align} x' &= (x^2+y^2)\sqrt{|y|}\\ y' &= \frac{1}{y^{42}+1}+\exp(t) \end{align}
and let $J \subseteq \mathbb{R}$ denote the maximal interval of existence for the solution satisfying $(x(0),y(0)) = (1,1)$.
I am asked to proof that $\sup J\leq 1$ and investigate how $(x(t),y(t))$ behaves if $t \nearrow \sup J$.
As a hint I am told to think about $z' = z^2$ for $z(0) = 1$ and look at the maximal interval of existence for this IVP. I did that and found the solution: $z(t) = \frac{1}{1-t}$. In this particular case we have $J = (-\infty,1)$ and $\lim_{t \to 1} z(t) = +\infty$.
Unfortunately I don't see how that might help for the initial problem. I have - after certainly thinking to long about this problem - not even the slightest idea about how to proof $\sup J \leq 1$.
How can this be done? I am thankful for any hint or solution sketch.
(I noticed that $x(t)$ and $y(t)$ are monotonically increasing.)
You can easily see $y'\ge e^t>0$ giving for $y\ge y(0)=1$ for $t\ge 0$ and thus $$ x'\ge x^2\implies x(t)\ge\frac1{1-t}. $$ Or a little better $$ x'\ge x^2+1\implies x(t)\ge\tan\left(\frac\pi4+t\right). $$