Is there a simple(computable) example of a smooth function $f$ with $f(0)=0$ and $f'(x)>0$ for all $x\in \Bbb R$, such that the solution of the ODE $$ y(0)=a, y'(x)=f(y)$$ cannot be defined for all $x\in \Bbb R$?
If the condition $f'(x)>0$ is replaced with $f'(x)\geq 0$ then $f(x)=x^3$ would be an example, but I can't find an example in the case $f'(x)>0$.
P.S. I need this result for a question about vector fields and flows
We can fairly easily make this happen for $a>0$ by requiring $y$ grow very fast. Choose $f(y)=e^y-1$ so that $$y'(x)=e^y-1.$$ You can check the solution to this with the condition $y(0)=a$ is $$y(x)=-\log(1+(e^{-a}-1)e^x).$$ One can then see this solution has a singularity at $$x=a-\log(e^a-1)$$ when $a>0$. For $a<0$, you can use $f(y)=e^{-y}-1$.