Concavity and inflection points of a Gaussian curve

First find $f'(x) = -2x e^{-x^2}$ using the chain rule. Differentiating again with the product rule gives $f''(x) = -2e^{-x^2} + (-2x)(-2x e^{-x^2}) = e^{-x^2}(4x^2 - 2)$. Since $e^{-x^2} > 0$ always, the sign of $f''(x)$ is determined entirely by the factor $4x^2 - 2$.

Setting $4x^2 - 2 = 0$ gives $x^2 = 1/2$, so $x = \pm 1/\sqrt{2}$. Testing values shows $f''(x) > 0$ (concave up) for $x < -1/\sqrt{2}$ and $x > 1/\sqrt{2}$, and $f''(x) < 0$ (concave down) for $-1/\sqrt{2} < x < 1/\sqrt{2}$, so concavity does change sign at both points. Thus both $x = \pm 1/\sqrt{2}$ are inflection points, with $f(\pm 1/\sqrt{2}) = e^{-1/2} = 1/\sqrt{e}$.