Taking the right-hand differential to calculate the slope of $xe^{-(\frac{1}{x}+ \frac{1}{|x|})}$ to the right of x=0, I got $e^{-\frac{2}{x}}(1 + \frac{2}{x})$, after simplifying it to $xe^{-\frac{2}{x}}$. Every other solution I found was by using the first principle; $\frac{he^{-\frac{2}{h}}-0}{h}$, h~0. This simplifies to $e^{-\frac{2}{h}}$, which is 0 at x=0.
Why did my normal differential fail here? And either way, how do I solve the differential I got, because it's $0\cdot\infty$?
After letting $f(x)=xe^{-(\frac{1}{x}+ \frac{1}{|x|})}$, and $t=\frac{2}{x}$, it follows that $$\lim_{x\to 0^+}f'(x)=\lim_{x\to 0^+}(e^{-\frac{2}{x}}(1 + \frac{2}{x}))=\lim_{t\to +\infty}e^{-t}(1 + t)=\lim_{t\to +\infty}\frac{1 + t}{e^t}=0$$ which is equal to your evaluation of $f_+'(0)$ (after extending $f$ at $x=0$ with the value $0$).
It is interesting to note that $$f'_-(0)=\lim_{h\to 0^-}\frac{h}{h}=1\not=0$$ and therefore such function is not differentiable at $x=0$, that is $f$ has not a "proper" slope at $x=0$.