Let $f:\mathbb{R} \to \mathbb{R}$ be an a infinitely differentiable function except at zero such that $f{^\prime}{^\prime}>0$ when it is defined. We have then that $f$ is a strictly convex function in both $(0,\infty)$ and in $(-\infty,0)$.
Can we ensure that $f$ is a strictly convex function in the whole $\mathbb{R}$?
I also know that $f$ is even and that it is increasing in $(0,\infty)$ if it helps.
Yes, the fact that $f$ is even and increasing in $(0,\infty)$ make this true (otherwise there'd be counterexamples like $e^{-|x|}$.) A convenient characterization of convexity is: $f:\mathbb{R}\to\mathbb{R} $ is convex if and only if for every $a\in\mathbb{R}$ there exists $k\in\mathbb{R}$ such that $f(x)\ge f(a)+k(x-a)$ for all $x\in \mathbb{R}$. In other words, the graph of $f$ admits a supporting line at every point (which is simply the tangent line at the point of differentiability).
In your situation, one can take $k=0$ when $a=0$ and $k=f'(a)$ when $a\ne 0$.