Let $y(x)$ the unique solution of the Cauchy problem
\begin{cases} y'(x)=\sqrt{\ln(1+y(x)^2)} \\ y(0)=y_0 \end{cases}
i) Show that, for $y_0>0$, $y(x)$ is strictly increasing and convex.
ii) Show that $y(x)$ is globally defined.
For i), I note that $$\ln(1+y(x)^2) = 0 \text{ iff } y(x)=0$$
So, if $y(0)=0$, then $y(x)=0$ is the unique solution.
If $y_0>0$, then in a right neighbourhood of $x=0$ I have that $y'(x)>0$. Also, by uniqueness, I can't cross the stationary solution $y(x)=0$, and therefore $$y'(x)=\ln(\sqrt{1+y^2})>0$$ for every $x>0$. Since this is the rhs of the equation, this proves that the function is strictly increasing.
To show convexity, I just computed $$y''(x)= \frac{y y'(x)}{1+y(x)^2} >0 $$ since $y'(x)>0$ and $y(x)>0$ by uniqueness.
ii) Here I don't know how to argue formally. It is clear from a picture that $\lim_{x \rightarrow \infty} y(x)=+\infty$, but I don't know how to show that $y(x)$ is globally defined.
Let $f(x,y) := \sqrt{\log(1 + y^2)}$ be the RH side of your ODE.
Since $\log$ is concave, you have $\log(1+t) \leq t$ for every $t >-1$ therefore:
$\log(1 + y^2) \leq y^2\quad \Rightarrow\quad f(x,y) \leq \sqrt{y^2} = |y|$
and $f(x,y)$ is sublinear in $y$ (uniformly w.r.t. $x$).
Now classical global existence theorems apply and ii follows.