I tried to go about about this using the definition : let $\phi$ be a test function
$$\begin{align}\langle \delta''(\frac13x^3 +x) ,\phi\rangle &= \langle \delta''(\frac13x^3 +x) ,\frac{x^2+1}{x^2+1}\phi\rangle = \langle (x^2+1)\delta''(\frac13x^3 +x) ,\frac{1}{x^2+1}\phi\rangle \\ & = \langle (\delta'(\frac13x^3 +x))' ,\frac{1}{x^2+1}\phi\rangle = - \langle \delta'(\frac13x^3 +x) ,\frac{1}{x^2+1}\phi' - \frac{2x}{(x^2+1)^2}\phi\rangle \\ & = - \langle (\delta(\frac13x^3 +x))' ,\frac{1}{(x^2+1)^2}\phi' - \frac{2x}{(x^2+1)^3}\phi\rangle \\ & = \langle \delta (\frac13x^3 +x) ,\frac{1}{(x^2+1)^2}\phi'' - \frac{2x-4x(x^2+1)^2}{(x^2+1)^3}\phi' - \frac{2(x^2+1)^3-6x(x^2+1)^2}{(x^2+1)^3}\phi\rangle \end{align} $$
now using that $$\langle \delta(F(x)) ,\phi\rangle = \sum_{x_i \in \ker F} \frac{\phi(x_i)}{|F'(x_i)|} $$
$F'(x) = x^2+1,\,x_i =0$
then
$$\langle \delta''(\frac13x^3 +x) ,\phi\rangle = \phi''(0)-2\phi(0) = \langle \delta''(x) -2 \delta(x) ,\phi\rangle$$
question number 1 : did I do anything not permitted ?
question number 2 : is there some some other ways to solve this ? maybe with less computations ?