Answer
Saddle Point of $f(0,0)=0$
Local Maximum $f(\dfrac{-1}{2},\dfrac{-1}{2} )=\dfrac{1}{4}$
Work Step by Step
We have
$f_x(x,y)=6x^2+3y=0; f_y(x,y)=3x+6y^2=0$
We solve the above equations for the critical points:
$(\dfrac{-1}{2},\dfrac{-1}{2} )$ and $(0,0)$
We will take the help of second derivative test, for $(\dfrac{-1}{2},\dfrac{-1}{2} )$ as follows:
$D=f_{xx}f_{yy}-f^2_{xy}=(0)(0)-(3)^2=-9 \lt 0$
So, Saddle Point of $f(0,0)=0$
We will take the help of second derivative test, for $(\dfrac{-1}{2},\dfrac{-1}{2} )$ as follows:
$D=f_{xx}f_{yy}-f^2_{xy}=(-6)(-6)-(3)^2=27 \gt 0$
So, Local Maximum $f(\dfrac{-1}{2},\dfrac{-1}{2} )=\dfrac{1}{4}$
