 Solution

At $-a$, the squared prefactor tends to $4$, so the oscillatory discontinuity again remains. Near zero,It is unbounded near zero, despite its assigned zero value there, so it is neither continuous nor differentiable at zero.

At $a$, put $x=a+h$. ThenThus $f_2$ is continuous and differentiable at $x=\pi /2$, with $f_2^{\prime }(\pi /2)=0$. After translating the point to zero, the estimate shows how quadratic damping restores differentiability at the origin. The only discontinuities and nondifferentiability points are $x=-\pi /2$ and $x=0$; the function is smooth away from these two points and $\pi /2$. At $\pi /2$, its derivative exists but is not continuous.

Indeed, for $x\ne a$ near $a$, put $B(x)=(x-a{)}^2/x^2$. Then $f_2^{\prime }=B^{\prime }\sin (\tan x)+B{\sec }^{2}x\cos (\tan x)$. Here $B^{\prime }\to 0$ and $B{\sec }^{2}x\to 1/a^2$, so the derivative retains oscillations. Existence of the derivative at a point does not require continuity of nearby derivatives.