 Solution

Set $a=\pi /2$ for this part. In the interval $(-\pi ,\pi )$ the special points are $-a,0,a$. Elsewhere $f_0=\sin (\tan x)$ is a composition of smooth functions.

At zero, $\tan x=x+O(x^3)$ and therefore $\sin (\tan x)=x+O(x^3)$. Its assigned value is zero, so the function is continuous and differentiable there, with $f_0^{\prime }(0)=1$.

At either tangent pole $\pm a$, the bounded sine oscillates without a limit. For an explicit check at $a$, let $x_j=\arctan (2\pi j+\pi /2)$ and $u_j=\arctan (2\pi j+3\pi /2)$. Both approach $a$ from below, but the function values are respectively $1$ and $-1$. Negating these sequences gives the same obstruction at $-a$. Assigning the value zero at the poles cannot repair either limit.

Thus $f_0$ is discontinuous and nondifferentiable exactly at $x=\pm \pi /2$, and is smooth elsewhere in the specified interval.