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.)
\[
f(x, y)=8-3 x^{2}-2 y^{2}
\]
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 \(\boxed{8}\).