Answer
The global minimum is $f\left( {1,0} \right) = f\left( { - 1,0} \right) = - \frac{1}{{\rm{e}}}$ and the global maximum is $f\left( {0,1} \right) = f\left( {0, - 1} \right) = \frac{4}{{\rm{e}}}$.
Work Step by Step
We have $f\left( {x,y} \right) = \left( {4{y^2} - {x^2}} \right){{\rm{e}}^{ - {x^2} - {y^2}}}$ on the domain ${x^2} + {y^2} \le 2$.
Step 1. Find the critical points on the domain and evaluate $f$ at these points
The partial derivatives are
${f_x} = - 2x{{\rm{e}}^{ - {x^2} - {y^2}}} - 2x\left( {4{y^2} - {x^2}} \right){{\rm{e}}^{ - {x^2} - {y^2}}}$
$ = \left( {2{x^3} - 8x{y^2} - 2x} \right){{\rm{e}}^{ - {x^2} - {y^2}}}$
${f_y} = 8y{{\rm{e}}^{ - {x^2} - {y^2}}} - 2y\left( {4{y^2} - {x^2}} \right){{\rm{e}}^{ - {x^2} - {y^2}}}$
$ = \left( { - 8{y^3} + 2{x^2}y + 8y} \right){{\rm{e}}^{ - {x^2} - {y^2}}}$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$:
${f_x} = \left( {2{x^3} - 8x{y^2} - 2x} \right){{\rm{e}}^{ - {x^2} - {y^2}}} = 0$
${f_y} = \left( { - 8{y^3} + 2{x^2}y + 8y} \right){{\rm{e}}^{ - {x^2} - {y^2}}} = 0$
Since ${{\rm{e}}^{ - {x^2} - {y^2}}} \ne 0$, the above set of equations reduce to
$x\left( {{x^2} - 4{y^2} - 1} \right) = 0$
$y\left( { - 4{y^2} + {x^2} + 4} \right) = 0$
From the first equation we obtain
$x=0$, ${\ \ \ }$ ${x^2} - 4{y^2} - 1 = 0$
From the second equation we obtain
$y=0$, ${\ \ \ }$ $ - 4{y^2} + {x^2} + 4 = 0$
Substituting $x=0$ in $ - 4{y^2} + {x^2} + 4 = 0$ gives $y = \pm 1$. So, the solutions are $\left( {0,1} \right)$ and $\left( {0, - 1} \right)$.
Substituting $y=0$ in ${x^2} - 4{y^2} - 1 = 0$ gives $x = \pm 1$. So, the solutions are $\left( {1,0} \right)$ and $\left( { - 1,0} \right)$.
So, the critical points are $\left( {0,0} \right)$, $\left( {0,1} \right)$, $\left( {0, - 1} \right)$, $\left( {1,0} \right)$ and $\left( { - 1,0} \right)$. The extreme values of $f$ corresponding to these critical points are $f\left( {0,0} \right) = 0$, $f\left( {0,1} \right) = \frac{4}{{\rm{e}}}$, $f\left( {0, - 1} \right) = \frac{4}{{\rm{e}}}$, $f\left( {1,0} \right) = - \frac{1}{{\rm{e}}}$ and $f\left( { - 1,0} \right) = - \frac{1}{{\rm{e}}}$.
Step 2. Check the boundaries
The boundary is the circle: ${x^2} + {y^2} = 2$.
We restrict the function $f$ along the boundary and obtain
$g\left( x \right) = \left( {4\left( {2 - {x^2}} \right) - {x^2}} \right){{\rm{e}}^{ - 2}} = \frac{1}{{{{\rm{e}}^2}}}\left( {8 - 5{x^2}} \right)$
We find the extreme value of $g$ by solving the equation $g'\left( x \right) = 0$:
$g'\left( x \right) = - \frac{{10}}{{{{\rm{e}}^2}}}x = 0$
The critical point is at $x=0$. So, the extreme value is $g\left( 0 \right) = \frac{8}{{{{\rm{e}}^2}}}$ or $f\left( {0, \pm \sqrt 2 } \right) = \frac{8}{{{{\rm{e}}^2}}}$.
Since the $x$-interval is $ - \sqrt 2 \le x \le \sqrt 2 $, we have $g\left( {\sqrt 2 } \right) = g\left( { - \sqrt 2 } \right) = - \frac{2}{{{{\rm{e}}^2}}}$.
Step 3. Compare the results
Comparing the results from Step 1 and Step 2 we obtain the global minimum $f\left( {1,0} \right) = f\left( { - 1,0} \right) = - \frac{1}{{\rm{e}}}$ and the global maximum $f\left( {0,1} \right) = f\left( {0, - 1} \right) = \frac{4}{{\rm{e}}}$.
