Find the critical point of the function. Then use the second derivative test to classify the nature of this point, if possible. (If an answer does not exist, enter DNE.)
$$f(x,y)=8-3x^2-2y^2$$
critical point $(x,y)=$ classification
relative maximum $\mathbf{0}$
Finally, determine the relative extrema of the function. (If an answer does not exist, enter DNE.)
relative minimum value
relative maximum value

Step 1 ：First, we find the critical points of the function by setting the first derivatives equal to zero. The first derivatives of the function are $f_x=-6x$ and $f_y=-4y$.

Step 2 ：Setting $f_x=0$ gives $x=0$, and setting $f_y=0$ gives $y=0$. So the critical point is $(0,0)$.

Step 3 ：Next, we use the second derivative test to classify the nature of this point. The second derivatives of the function are $f_{xx}=-6$, $f_{yy}=-4$, and $f_{xy}=0$.

Step 4 ：The determinant of the Hessian matrix is $D=f_{xx}{f}_{yy}-(f_{xy}{)}^2=(-6)(-4)-0=24$. Since $D>0$ and $f_{xx}<0$, the point $(0,0)$ is a relative maximum.

Step 5 ：Finally, we determine the relative extrema of the function. The function value at the critical point $(0,0)$ is $f(0,0)=8-3(0{)}^2-2(0{)}^2=8$.

Step 6 ：So, the relative minimum value does not exist (DNE), and the relative maximum value is $\boxed{{\displaystyle 8}}$.