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