$R{f}^{\text{'}}\left(0\right)=\underset{h\to 0}{\lim }\frac{e^{-\frac{1}{{\left(0+h\right)}^2}}-0}{h}=\underset{h\to 0}{\lim }\frac{1}{h}{e}^{-\frac{1}{h^2}}$
$=\underset{h\to 0}{\lim }\frac{\frac{1}{h}}{e^{\frac{1}{h^2}}}$
$=\underset{h\to 0}{\lim }\frac{-\frac{1}{h^2}}{e^{\frac{1}{h^2}}\cdot \left(-\frac{2}{h^3}\right)}$
$=\underset{h\to 0}{\lim }\frac{h}{2e^{\frac{1}{h^2}}}=\frac{0}{\infty }=0$
$L{f}^{\text{'}}\left(0\right)=\underset{h\to 0}{\lim }\frac{e^{-\frac{1}{{\left(0-h\right)}^2}}-0}{-h}=\underset{h\to 0}{\lim }\left(-\frac{1}{h}\right)e^{-\frac{1}{h^2}}$
$=\underset{h\to 0}{\lim }\frac{-\frac{1}{h}}{e^{\frac{1}{h^2}}}$
$=\underset{h\to 0}{\lim }\frac{\frac{1}{h^2}}{e^{\frac{1}{h^2}}\left(-\frac{2}{h^3}\right)}$
$=\underset{h\to 0}{\lim }\frac{h}{-2e^{\frac{1}{h^2}}}=\frac{0}{\infty }=0$

Hence, $f^{\text{'}}\left(0\right)=0$