Answer
Relative maximum at (-1,0)
Work Step by Step
$-2(1+1 x) \cdot e^{-\left(x^{2}+y^{2}+2 x\right)}=f_{x}(x, y)$
$-2 y e^{-\left(x^{2}+y^{2}+2 x\right)}=f_{y}(x, y)$
At the critical point.
$-2(1+1 x) \cdot e^{-\left(x^{2}+y^{2}+2 x\right)}=0$
$-2 y \cdot e^{-\left(x^{2}+y^{2}+2 x\right)}=0$
so, $x=-1, y=0$
So, it follows:
$-2 e^{-\left(x^{2}+y^{2}+2 x\right)}+4(1+x)^{2} \cdot e^{-\left(x^{2}+y^{2}+2 x\right)}=f_{x x}(x, y)$
$-2 e^{-\left(x^{2}+y^{2}+2 x\right)}+4 y^{2} \cdot e^{-\left(x^{2}+y^{2}+2 x\right)}=f_{y z}(x, y)$
$4(1+x) \cdot y \cdot e^{-\left(x^{2}+y^{2}+2 x\right)}=f_{x x}(x, y)$
$-2 e=f_{x x}(-1,0)$
$-2 e=f_{y y}(-1,0)$
$0=f_{x x}(-1,0)$
\[
D(-1,0)=4 \mathrm{e}^{2}>0, \text { and } f_{x x}(-1,0)<0
\]
Relative maximum at (-1,0)
