Let [ ] denotes the greatest integer function and $f\left(x\right)=\left[{\tan }^{2}x\right]$, then

Here, $\left[.\right]$ denotes the greatest integral function.

Thus, if $-45°<x<45°$

$$\begin{gathered} \Rightarrow \tan \left(-45°\right)<\tan x<\tan \left(45°\right) \\ \Rightarrow -1<\tan x<1 \\ \Rightarrow 0<{\tan }^{2}x<1 \end{gathered}$$

Since, $f\left(x\right)=\left[{\tan }^{2}x\right]$ so $f\left(x\right)=0 \forall x\in \left[-\frac{\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{4}\right]$

Therefore, $f\left(x\right)$ is zero for all values of $x$ from $\left(-45°\right)$ to$\left(45°\right)$. 

Thus, $f\left(x\right)$ exists when $x\to 0$ and also it is continuous at $x=0$

$f\left(x\right)$ is differentiable at $x=0$ and has a value $0$. (as $f\left(x\right)=0$ ).

Hence, option-$2$ is the correct answer.