Answer
$f(\dfrac{1}{2},0)$ is a local maximum and $f(\dfrac{-1}{2},0)$ local minimum.
Work Step by Step
Second derivative test: Some noteworthy points to calculate the local minimum, local maximum and saddle point of $f$.
1. If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\gt 0$ , then $f(p,q)$ is a local minimum.
2.If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\lt 0$ , then $f(p,q)$ is a local maximum.
3. If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \lt 0$ , then $f(p,q)$ is not a local minimum, local maximum, or a saddle point.
For the given question, we have $f_x(x,y)=e^{-2x^2-2y^2}(1-4x^2)$ and $f_y(x,y)=-4xye^{-2x^2-2y^2}$ gives $(x,y)=(\dfrac{1}{2},0),(\dfrac{-1}{2},0)$.
For $(x,y)=(\dfrac{1}{2},0)$
$D(\dfrac{1}{2},0)=8e^{-1} \gt 0$ and $f_{xx}(\dfrac{1}{2},0)=e^{-1/2}(-4) \lt 0$
For $(x,y)=(\dfrac{1}{2},0)$
$D(\dfrac{-1}{2},0)=8e^{-1} \gt 0$ and $f_{xx}(\dfrac{1}{2},0)=e^{-1/2}(4) \gt 0$
This implies that $f(\dfrac{1}{2},0)$ is a local maximum and $f(\dfrac{-1}{2},0)$ local minimum.
