Suppose f(x)=$e^{-\sqrt{x}}+e^{-1/x^2}.If f\text{'}\text{'}\left(x\right)=\alpha \frac{e^{-\sqrt{x}}}{x}\left(1+\frac{1}{\sqrt{x}}\right)+\beta \frac{e^{1/x^2}}{x^4}\left(3-\frac{2}{x^2}\right), then (\alpha ,\beta )=$

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{-\sqrt{\mathrm{x}}}+{\mathrm{e}}^{-\frac{1}{{\mathrm{x}}^2}} \\ {\mathrm{f}}^1\left(\mathrm{x}\right)={\mathrm{e}}^{-\sqrt{\mathrm{x}}}\left(-\frac{1}{2\sqrt{\mathrm{x}}}\right)+{\mathrm{e}}^{-\frac{1}{{\mathrm{x}}^2}}\left(+\frac{2}{{\mathrm{x}}^3}\right) \\ {\mathrm{f}}^{11}\left(\mathrm{x}\right)={\mathrm{e}}^{-\sqrt{\mathrm{x}}}{\left(\frac{-1}{2\sqrt{\mathrm{x}}}\right)}^2+\frac{1}{4}\frac{1}{{\mathrm{x}}^{{\displaystyle \frac{3}{2}}}}{\mathrm{e}}^{-\sqrt{\mathrm{x}}}+{\mathrm{e}}^{\frac{-1}{{\mathrm{x}}^2}}{\left(\frac{-2}{{\mathrm{x}}^3}\right)}^2-\frac{6}{{\mathrm{x}}^4}{\mathrm{e}}^{\frac{-1}{{\mathrm{x}}^2}} \\ =\frac{1}{4\mathrm{x}}{\mathrm{e}}^{-\sqrt{\mathrm{x}}}+\frac{1}{4\mathrm{x}\sqrt{\mathrm{x}}}{\mathrm{e}}^{-\sqrt{\mathrm{x}}}+{\mathrm{e}}^{\frac{-1}{{\mathrm{x}}^2}}\frac{4}{{\mathrm{x}}^6}-\frac{6}{{\mathrm{x}}^4}{\mathrm{e}}^{\frac{-1}{{\mathrm{x}}^2}} \\ =\frac{1}{4}\frac{{\mathrm{e}}^{-\sqrt{\mathrm{x}}}}{\mathrm{x}}\left(1+\frac{1}{\mathrm{x}}\right)-\frac{2}{{\mathrm{x}}^4}{\mathrm{e}}^{\frac{-1}{{\mathrm{x}}^2}}\left(3-\frac{2}{{\mathrm{x}}^2}\right) \\ \mathrm{\alpha }=\frac{1}{4}, \mathrm{\beta }=-2 \end{gathered}$$