Answer
The minimum value of $f$ is $\frac{1}{{\rm{e}}}$ and the maximum value of $f$ is $1$.
Work Step by Step
We have $f\left( {x,y} \right) = {{\rm{e}}^{ - {x^2} - {y^2}}} = \frac{1}{{{{\rm{e}}^{\left( {{x^2} + {y^2}} \right)}}}}$.
The domain of $f$ is given to satisfy ${x^2} + {y^2} \le 1$.
Since ${x^2} + {y^2} \ge 0$, we can write $0 \le {x^2} + {y^2} \le 1$.
Since ${{\rm{e}}^{\left( {{x^2} + {y^2}} \right)}} > 0$ and is increasing, the minimum of $f$ occurs when ${{\rm{e}}^{\left( {{x^2} + {y^2}} \right)}}$ is maximum, that is when ${x^2} + {y^2} = 1$ and the maximum of $f$ occurs when ${{\rm{e}}^{\left( {{x^2} + {y^2}} \right)}}$ is minimum, that is at $\left( {0,0} \right)$. Thus, the minimum value of $f$ is $\frac{1}{{\rm{e}}}$ and the maximum value of $f$ is $1$.
