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.
Best Answer
We can assume by symmetry and without loss of generality that the ellipse can be parametrized by $$(x,y) = (a \cos \theta, b \sin \theta), \quad a, b > 0, \quad \theta \in [0,2\pi).$$ We require tangency to the curve $y = e^{-x^2}$ as well as a single point of intersection in the first quadrant. That is to say, $$b \sin \theta = e^{-(a \cos \theta)^2}$$ has a unique solution for $\theta \in (0, \pi/2)$, and at this point, $$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = -\frac{b}{a} \cot \theta = -2x(\theta)e^{-x(\theta)^2} = -2a (\cos \theta )e^{-(a \cos \theta)^2}.$$ Consequently, $$-\frac{b}{a} \cot \theta = -2ab \cos \theta \sin \theta,$$ or $$\sin \theta = \frac{1}{a \sqrt{2}}.$$ Note if $a < 1/\sqrt{2}$, no such angle exists. The ellipse is "too narrow"--this is due to the fact that the point of tangency is at $(x,y) = (0,1)$. At the point of tangency, we also have $$\cos \theta = \sqrt{1 - (2a^2)^{-1}}$$ so that we now have $$\frac{b}{a \sqrt{2}} = e^{1/2 - a^2}$$ or $$b = a e^{1/2 - a^2} \sqrt{2}.$$ Finally, we seek to maximize the area of this family of ellipses parametrized by $a$. Since the area is proportional to $ab$, we need to maximize $$f(a) = a^2 e^{1/2-a^2}.$$ Computing the derivative with respect to $a$ and solving for critical values, we get $$0 = \frac{df}{da} = 2(a-1)a(a+1)e^{1/2-a^2},$$ hence $$a = 1$$ is the unique solution, with $b = \sqrt{2/e}$ and the ellipse has equation $$\frac{x^2}{1} + \frac{y^2}{2/e} = 1.$$ The area of this ellipse is simply $$\pi a b = \pi \sqrt{\frac{2}{e}}.$$ This is obviously not a circle.
For your enjoyment, I have included an animation of the family of ellipses $$\frac{x^2}{a^2} + \frac{y^2}{2a^2 e^{1-2a^2}} = 1,$$ for $a \in [1/\sqrt{2},2]$: