Answer
See explanation below for the proof.
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 and local maximum or, a saddle point.
As we are given that $f(x,y)=x^2ye^{-x^2-y^2}$
For $(x,y)=( 1,\dfrac{1}{\sqrt 2})$
$D=8e^{-3} \gt 0$ ; and $f_{xx} \lt 0$
For $(x,y)=( -1,\dfrac{1}{\sqrt 2})$
$D=8e^{-3} \gt 0$ ; and $f_{xx} \lt 0$
For $(x,y)=( 1,-\dfrac{1}{\sqrt 2})$
$D=8e^{-3} \gt 0$ ; and $f_{xx} \gt 0$
For $(x,y)=( -1,-\dfrac{1}{\sqrt 2})$
$D=8e^{-3} \gt 0$ ; and $f_{xx} \gt 0$
$D(0,0) \lt 0$; saddle points; because when $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \lt 0$ , then $f(p,q)$ is not a local minimum and local maximum or, a saddle point.
Also, $f_{xx}=2y,f_{yy}=0, f_{xy}=0$
and $f_{xx}(x,y)f_{yy}(x,y)-[f_{xy}(x,y)]^2 = 0$
Therefore, we have Maximum value: $f(\pm 1, \dfrac{1}{\sqrt 2})$, Minimum value: $f(\pm 1, -\dfrac{1}{\sqrt 2})$ and saddle point:$(0, 0)$.
This implies that $D=0$ that is, the function $f$ has infinite critical points along the line $x=0$ on the line $x=0$ at the points $f(\pm 1, \pm \dfrac{1}{\sqrt 2})$
Hence, the result has been proved.
