Concavity of tan(x) on (-π/2, π/2)

Since $\sec^2(x) > 0$ for all $x$ in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, the sign of $f''(x) = 2\sec^2(x)\tan(x)$ matches the sign of $\tan(x)$. On $\left(-\frac{\pi}{2}, 0\right)$, $\tan(x) < 0$, so $f''(x) < 0$, meaning $f$ is concave down there.

On $\left(0, \frac{\pi}{2}\right)$, $\tan(x) > 0$, so $f''(x) > 0$, meaning $f$ is concave up there. Since $f''$ changes sign from negative to positive as $x$ passes through $0$, and $f''(0) = 0$, the point $x = 0$ (where $f(0)=0$) is indeed an inflection point of $f$.