Prove or disprove that the ellipse of largest area (centered at origin) inscribed in $y=\pm e^{-x^2}$ has the equation $x^2+y^2=\frac12(1+\log2)$. I can show that $x^2+y^2=\frac12(1+\log2)$ is the equation of the circle of largest area inscribed in $y=\pm e^{-x^2}$:

The minimum distance $r$ (which will be the radius of the circle) between the origin and $y=e^{-x^2}$  can be found by finding the critical numbers of the derivative of the distance function.
  \begin{align}
r&=\sqrt{x^2+(e^{-x^2})^2}\\
\frac{dr}{dx}&=\frac{2x-4xe^{-2x^2}}{2\sqrt{x^2+e^{-2x^2}}}\\
0&=2x(1-2e^{-2x^2})\\
x&=0\quad\text{(obviously inadmissible, or)}\\
x&=\sqrt{\frac12\log2}\\
\\
r&=\sqrt{\frac12\log2+e^{-\log2}}\\
r^2&=\frac12\log2+\frac12\\
\implies\quad x^2+y^2&=\frac12\left(1+\log2\right)
\end{align}
  as desired.

The relationship between a circle and an ellipse is like that of a square and a rectangle:  given a set perimeter, the more square the rectangle, the larger the area.  But since this question is based on the curve $y=e^{-x^2}$ and not on any set perimeter, it does not seem like the same conclusion can be drawn.  I would need something stronger (perhaps based on concavity?) to show whether $x^2+y^2=\frac12(1+\log2)$ is the largest ellipse or not.

$$
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \implies 
\frac{dy}{dx}=-\frac{b^2}{a^2}\frac xy=-\frac{b}{a^2}\frac x{\sqrt{1-\frac{x^2}{a^2}}}.
$$
Hence we have the following system to find the tangent point:
$$\begin{cases}
e^{-x^2}=b\sqrt{1-\frac{x^2}{a^2}}\\
2xe^{-x^2}=\frac{b}{a^2}\frac x{\sqrt{1-\frac{x^2}{a^2}}}
\end{cases}\implies
1-\frac{x^2}{a^2}=\frac{1}{2a^2}\implies x^2=a^2-\frac12.\tag1
$$
Substituting this back into the equation of the tangent point one obtains equation to determine $b$:
$$
e^{\frac12-a^2}=\frac ba\sqrt{\frac12}\implies b=\sqrt{2e}\,ae^{-a^2}.
$$
The area is respectively:
$$
A=\pi ab=\pi \sqrt{2e}\,a^2e^{-a^2}.\tag2
$$
To find the extremum of the area we differentiate over $a$ to obtain:
$$
\frac1{\pi\sqrt{2e}}\frac{dA}{da}=2ae^{-a^2}-2a^3e^{-a^2}=2ae^{-a^2}(1-a^2),
$$
meaning that the largest value of area is achieved at $a=1$:
$$
A_\text{max}=\pi\sqrt{\frac2e}.
$$
Since $a\ne b$ the ellipse of the largest area is not a circle.

PS. In fact the system (1) has another solution: $x=0, b=1$. It can be however shown that the largest possible area of the inscribed ellipse in this case is $\frac\pi{\sqrt2}$ which corresponds to $a=\frac1{\sqrt2}$ in equation (2).
