Answer
Saddle point at $(0,0)$; Local maximum at $(\dfrac{-2}{3},\dfrac{2}{3})=\dfrac{170}{27}$
Work Step by Step
Since, we have $f_x(x,y)=3x^2-2y=0, f_y(x,y)=-3y^2-2x=0$
After solving the above two equations, we get
The critical point is: $(0,0)$ and $(\dfrac{-2}{3},\dfrac{2}{3})$
As per second derivative test, for $(0,0)$ we have
$D=f_{xx}f_{yy}-f^2_{xy}=-4$ and $D=-4 \lt 0$
Thus, saddle point at $(0,0)$
Now, for $(\dfrac{-2}{3},\dfrac{2}{3})$, we have
$D=f_{xx}f_{yy}-f^2_{xy}=12$ and $D=12 \gt 0$ and $f_{xx}=-4 \lt 0$
Thus, local maximum at $(\dfrac{-2}{3},\dfrac{2}{3})=\dfrac{170}{27}$
Hence: Saddle point at $(0,0)$; Local maximum at $(\dfrac{-2}{3},\dfrac{2}{3})=\dfrac{170}{27}$
