解：

\(\frac{\partial f}{\partial x} = -2x\mathrm{e}^{-y} \Rightarrow f(x,y) = -x^2\mathrm{e}^{-y} + \varphi(y) \)。则\(\frac{\partial f}{\partial y} = x^2\mathrm{e}^{-y} + \varphi^\prime(y) = \mathrm{e}^{-y}x^2 - (y + 1)\mathrm{e}^{-y}\)

则\(\varphi^\prime(y) = -(y + 1)\mathrm{e}^{-y}\)

\(\Rightarrow \varphi(y) = (y + 2)\mathrm{e}^{-y} + C\)

则

\(f(x,y) = -x^2\mathrm{e}^{-y} + (y + 2)\mathrm{e}^{-y} + C\)

又\(f(0,0) = 2\)，\(\Rightarrow C = 0\)

则\(f(x,y) = -x^2\mathrm{e}^{-y} + (y + 2)\mathrm{e}^{-y} \)。

\(\begin{cases}

\frac{\partial f}{\partial x} = -2x\mathrm{e}^{-y} = 0 \\

\frac{\partial f}{\partial y} = \mathrm{e}^{-y}(x^2 - y - 1) = 0

\end{cases} \Rightarrow \begin{cases}

x = 0 \\

y = -1

\end{cases}\)

则驻点\((0, -1)\)

\(f_{xx} = -2\mathrm{e}^{-y}, f_{xy} = 2x\mathrm{e}^{-y}, f_{yy} = \mathrm{e}^{-y}(x^2 - y - 1) - \mathrm{e}^{-y} = \mathrm{e}^{-y}(x^2 - y) \)

在点\((0, -1)\)处\(A = -2\mathrm{e}, B = 0, C = -\mathrm{e}\)

则\(AC - B^2 > 0, A < 0\)，从而\( f(x,y) \)在\((0, -1)\)处有极大值，且极大值为

\(f(0, -1) = \mathrm{e}\)