Answer
Saddle Point of $f(0,0)=15$
and
Local Maximum $f(1,1 )=14$
Work Step by Step
$$f_x(x,y)=3x^2-3y=0 $$ and $$f_y(x,y)=3y^2-3x=0$$
Simplify the above two equations to get the critical points.
Thus, the critical point are: $(1,1 )$ and $(0,0)$
Apply the second derivative test for $(0,0)$.
$D=f_{xx}f_{yy}-f^2_{xy}=(0)(0)-(-3)^2=0-9=-9 \lt 0$
Saddle Point of $f(0,0)=15$
Apply the second derivative test for $(1,1)$.
$D=f_{xx}f_{yy}-f^2_{xy}=(6)(6)-(-3)^2=36-9=27 \gt 0$
Local Maximum of $f(1,1 )=14$
