why here $x^2$ is used ? why not $x ?$ 
Does  there exists a function $f : \mathbb{R }\rightarrow \mathbb{R}$ which is differentiable only at the point $0.$?

My attempt :  I found the  answer  here Is there a function $f: \mathbb R \to \mathbb R$ that has only one point differentiable?
But  i didn't understands the  answer , my doubts given below

 A: Because while $x p(x)$ is continuous at $0$, it is not differentiable.
In particular, the fraction
$$
\frac{(0+h)p(0+h)-0p(0)}{h}
$$
has value $0$ or $1$ depending on whether $h$ is rational or not. So it has no limit as $h\to 0$, which by definition of derivative means that $xp(x)$ had no derivative at $0$.
On the other hand, the fraction
$$
\frac{(0+h)^2p(0+h)-0^2p(0)}{h}
$$
has value $h$ or $0$ depending on whether $h$ is rational or irrational. Thus it does have a limit as $h\to0$, which is to say that $x^2p(x)$ has derivative $0$ at $x=0$.
A: The function with $x$ is not differentiable at $0$. For instance in the limit defining the derivative, the $\lim \sup$ exists and is $1$, and the $\lim \inf$ exists and is $0.$ These are not equal, so the derivative doesn't exist at $0.$
A: Note that $f(0)=0$ and $\frac{f(x)}{x}=xp(x)$
And the limit of $xp(x)$ at zero is zero.
If $f=xp(x)$ then $\frac{f(x)}{x}=p(x)$
and the limit of $p(x)$ at zero does not exist.
A: With $f(x)=xp(x),$ the derivative at $0$ would be $\lim\limits_{x\to0}\dfrac{f(x)-f(0)}{x-0}=\lim\limits_{x\to0}\dfrac{xp(x)}x=\lim\limits_{x\to0}p(x),$ 
which does not exist, since $p(x)$ takes values of $0$ and $1$ for arbitrarily small $x$.
