By subdifferential I mean the set of all subderivatives, where the subderivative at point $x$ is a real number $c$ such that for all $y \in \mathbb{R}$ it is the case that: $$f(y) - f(x) \geq c(y-x)$$
It seems to me that for $x < 0$ it is just $0$ and for $x > 0$ it is $2x\exp(x^2)$. So I am interested in subdifferential at point $x = 0$.
So I have to find $c$ such that for all $y$ $$f(y) - f(x) \geq c(y -x) \\ f(y) - 1 \geq cy \\$$
$f(y)$ has to be greater or equal to $1$.
Is there other subdifferential at point $x = 0$ then $c = 0$?
In general at a breakpoint the subgradient is the convex hull of the left and right derivative. In this case you have $f(y) - 1 \geq cy$ for all $y \in \mathbb{R}$, so \begin{align} 0 \geq cy \qquad \forall y \leq 0 \\ \exp(y^2) - 1 \geq cy \qquad \forall y \geq 0 \end{align} The first inequality gives $c \geq 0$. The second inequality gives $c \leq [\exp(y^2) - \exp(0)] / y$, which if you take the limit of $y \downarrow 0$ gives $c \leq 0$, so $c=0$.