Find the local extrema for $f(x, y)=x^{3}+y^{3}+6 x y+1$ A. $f(2,2)=41$ is a local maximum B. $f(0,0)=1$ is a local minimum C. $f(-2,-2)=9$ is a local maximum D. $f(1,1)=-1$ is a local minimum

Step 1 ：Find the first derivatives of the function with respect to x and y. The first derivative with respect to x is \(f_x = 3x^2 + 6y\) and with respect to y is \(f_y = 3y^2 + 6x\).

Step 2 ：Set the first derivatives equal to zero to find the critical points. This gives the equations \(x^2 + 2y = 0\) and \(y^2 + 2x = 0\).

Step 3 ：Solve these equations simultaneously to find the critical points. The solutions are \((x, y) = (0, 0)\) and \((x, y) = (-2, -2)\).

Step 4 ：Compute the second derivatives to form the Hessian matrix. The second derivatives are \(f_{xx} = 6x\), \(f_{yy} = 6y\), and \(f_{xy} = f_{yx} = 6\).

Step 5 ：For the point (0, 0), the Hessian matrix is \[H = \begin{bmatrix} 0 & 6 \\ 6 & 0 \end{bmatrix}\]. The determinant of this matrix is \(0 - 36 = -36\), which is less than 0, so (0, 0) is a saddle point, not a local extremum.

Step 6 ：For the point (-2, -2), the Hessian matrix is \[H = \begin{bmatrix} -12 & 6 \\ 6 & -12 \end{bmatrix}\]. The determinant of this matrix is \((-12)*(-12) - 6*6 = 144 - 36 = 108\), which is greater than 0. Since both \(f_{xx}\) and \(f_{yy}\) are negative, (-2, -2) is a local maximum.

Step 7 ：Therefore, the function \(f(-2,-2)=9\) is a local maximum. So, the correct answer is \(\boxed{9}\).