Find max/min of $e^{2x}\left(x+y^{2}+2y\right)$ $$e^{2x}\left(x+y^{2}+2y\right)$$
FOC: $$\begin{cases}
2e^{2x}\left(x+y^{2}+2y\right)+e^{2x}=0\\
e^{2x}\left(2y+2\right)=0
\end{cases}\rightarrow\begin{cases}
x=\frac{1}{2}\\
y=-1
\end{cases}$$
SOC: $$\begin{pmatrix}4e^{2x}\left(x+y^{2}+2y\right)+2e^{2x} & 2e^{2x}\left(2y+2\right)\\
2e^{2x}\left(2y+2\right) & 2e^{2x}
\end{pmatrix}\rightarrow\begin{pmatrix}0 & 0\\
0 & 2e
\end{pmatrix}$$
As both leading principal minors are equal to 0, Hessian is positive semidefinite but not positive definite, i.e. the function is convex but not strictly. So I cannot rule out the possibility that the function has local minimum at this point.
WolframAlpha tells that the point, in fact, is local minimum. How can I get such result?
 A: Note that it is a global minimum:
$$f(x,y)=e^{2x}(x+y^2+2y)=e^{2x}(y^2+2y+1+x-1)=e^{2x}((y+1)^2+x-1)\geq e^{2x}(x-1)=g(x)\geq -\frac{e}{2}=f(\frac12,-1).$$
So, it only remains to show that $g(x)\geq -e/2,$ which is quite easy.
A: Your second partial derivative relatively to $x$ is false : $2e^{2x}(2x+2y^2+4y+1)+2e^{2x}=2e$
$\begin{pmatrix}2e & 0\\0 & 2e\end{pmatrix}$ Hessian is positive i.e. the function is strictly convex. Hence  the point is local minimum.
A: Let's
$$
u(x,y)=e^{2x}\left(x+y^{2}+2y\right).
$$
So we have
$$
u(x+\Delta x,y+\Delta y)=e^{2x+2\Delta x}\left(x+\Delta x+(y+\Delta y)^{2}+2y+2\Delta y\right).
$$
And
$$
u(x+\Delta x,y+\Delta y)-u(x,y)=e^{2x}\left(e^{2\Delta x}(x+\Delta x)+e^{2\Delta x}(y+\Delta y)^{2}+2e^{2\Delta x}(y+\Delta y)-x-y^2-2y\right).
$$
so
$$
u(\frac{1}{2}+\Delta x,-1+\Delta y)-u(\frac{1}{2},-1)=e\left(e^{2\Delta x}(\frac{1}{2}+\Delta x)+e^{2\Delta x}(-1+\Delta y)^{2}+2e^{2\Delta x}(-1+\Delta y)+\frac{1}{2}\right)\ge e^{2\Delta x}\left(\Delta x + \Delta y^2-\frac{1}{2}\right)+\frac{1}{2}\ge e^{2\Delta x}\left(\Delta x-\frac{1}{2}\right)+\frac{1}{2}\ge0
$$
