Question 1035337
{{{f(x,y)[x] = 3x^2+6y}}} and {{{f(x,y)[y] = 3y^2+6x}}}.

Also, {{{f[xx] = 6x}}} and {{{f[yy] = 6y}}}, and {{{f[xy]= f[yx] = 6}}}

Use the Second partial derivative test.

Set the partial derivatives above to 0 and solve for x and y.

{{{3x^2+6y = 0}}} and {{{ 3y^2+6x = 0}}}.

This system of equations yield two points as its solutions:  (0,0) , (-2,-2).  (Verify!!)

Now find the determinant of the hessian for each point.

For (0,0): {{{f[xx]*f[yy] - (f[xy])^2 = 0*0-6^2 = -36 < 0}}}.

This implies that the point (0,0) is a saddle point (neither a local max nor a local min).

For (-2,-2):  {{{f[xx]*f[yy] - (f[xy])^2 = -12*-12-6^2 = 144-36 > 0}}}.
Since {{{f[xx] < 0}}}, it follows that (-2,-2) is a local maximum.