Let $f:R\to R$ satisfying $\left|f\right(x\left)\right|\le x^2,\forall x\in R$, then $f\left(x\right)$ is satisfying which of the following for sure?

Given: $\left|f\right(x\left)\right|\le x^2,\forall x\in R$.

Therefore,

At $x=0, \Rightarrow \left|f\right(0\left)\right|\le 0$

$\Rightarrow f\left(0\right)=0$.

So,

$$\begin{gathered} \Rightarrow f^{\text{'}}\left(0\right)=\underset{h\to 0}{\lim }\left(\frac{f\left(h\right)-f\left(0\right)}{h}\right) \\ \Rightarrow f^{\text{'}}\left(0\right)=\underset{h\to 0}{\lim }\left(\frac{f\left(h\right)}{h}\right) - \left(i\right) \end{gathered}$$

Now, 

$\Rightarrow \left|\frac{f\left(h\right)}{h}\right|\le \left|h\right|$

$\Rightarrow -\left|h\right|\le \frac{f\left(h\right)}{h}\le \left|h\right|$

Using Cauchy-Squeeze theorem:

$\Rightarrow \underset{h\to 0}{\lim }\left(\frac{f\left(h\right)}{h}\right)\to 0 - \left(ii\right)$

From Eqs. $\left(i\right)$ and $\left(ii\right),$ we get:

$\Rightarrow f^{\text{'}}\left(0\right)=0$

Therefore, $f\left(x\right)$ is differentiable at $x=0.$

So the correct option is $\left(1\right).$