Derivative with an absolute value

For $x>0$: $f(x)=x+e^x$, so $f'(x)=1+e^x$.

For $x<0$: $f(x)=-x+e^x$, so $f'(x)=-1+e^x$.

At $0$. The difference quotient is $\dfrac{|h|+e^h-1}h=\dfrac{|h|}h+\dfrac{e^h-1}h$. The second term tends to $1$. The first term is $+1$ for $h>0$ and $-1$ for $h<0$. So the right derivative is $1+1=2$ and the left derivative is $-1+1=0$. They differ, so $f$ is not differentiable at $0$.

Answer. $f$ is differentiable on $\mathbb{R}^*$, with $f'(x)=e^x+1$ for $x>0$ and $f'(x)=e^x-1$ for $x<0$.

Remark. The sum of a differentiable function and a function with a corner has a corner at the same place: the exponential only changes the slopes on both sides.