Differentiability of $f(x)$ and continuity of $f'(x)$: same thing or different? If a function $f(x)$ is differentiable $f'(a+)=f'(a-)$ at any point $a$ , so does it mean necessarily that $f'(x)$ is continuous at that point.
Also vice versa.. if $f'(x)$ is continuous at any point, does it mean $f(x)$ is differentiable at that point?
which of above two conclusion is valid ?
My main confusion is about following question -
$$
f(x)= 
\left\{ \begin{array}{lr} 
     x^2 \sin\left(\frac{1}{x}\right) & x>0 \\ 
     0 & x=0 \\
     x^2 \sin\left(\frac{1}{x}\right) & x < 0
    \end{array} \right.
$$
Here function is differentiable at $x=0$, but $f'(x)$ is not continuous at $x=0$ ...
 A: The derivative of a function doesn't need to be continuous. Your example is one counterexample. All the problems with the considered function appear at $0$, since that's where the formula $x^2\sin \frac{1}{x}$ does not work anymore.
For a start, notice that your function is continuous, since when you take $x\to 0$ you have zero multiplied by a bounded quantity ($\sin$ is bounded).
When the function is given like this, you can study the continuity in $0$ only by definition. 
$$\lim_{x \to 0} \frac{f(x)-f(0)}{x-0}=\lim_{x \to 0}x \sin \frac{1}{x}=0 $$
therefore $f'(0)$ exists and it equals zero.
For every other $x\neq 0$ you can differentiate like usual, using the formulas
$$\left( x^2\sin \frac{1}{x} \right)'=2x \sin \frac{1}{x}+x^2 \cos \frac{1}{x}\left(-\frac{1}{x^2}\right)=2x \sin\frac{1}{x}-\cos \frac{1}{x}$$
Notice that this function is not continuous in $0$. 
Now you have 
$$f'(x)=\begin{cases}0 & x=0\\  2x \sin\frac{1}{x}-\cos \frac{1}{x} & x \neq 0\end{cases}$$
a function which is not continuous.

Maybe you are asking for a criterion for differentiability of a function using a corollary of Lagrange's theorem. You can see my answer in this post for more details, but this does not apply to your case, since the limit $\lim_{x \to 0}f'(x)$ does not exist.
To answer the second part of your question, the corollary presented in the link says that if you can calculate $f'(x)$ around $x_0$ but not in $x_0$ (directly) and $\lim\limits_{x \to x_0} f'(x)$ exists and it is finite, then $f'(x_0)$ also exists and it equals the previous limit.
A: On the other hand, derivatives do have the intermediate value property; that is, if $f\colon [a, b] \to \mathbf R$ is differentiable and $f'(a) < d < f'(b)$ then there exists a $c \in (a, b)$ such that $f'(c) = d$. This is sometimes called Darboux's theorem.
It follows from this that the discontinuities of $f'$ cannot be "jumps" for which $f'(a\pm)$ both exist but are simply unequal. They have to be crazy (there might be a more precise term), as in your example.
A: I have the impression that you are mixing notations.
$$f'_+(a)=\lim_{x\to a^+}\frac{f(x)-f(a)}{x-a}$$
is the right derivative of $f$ at $x=a$. On the other hand, if $f$ is diferentiable on a neighbourhood of $a$,
$$
f'(a+)=\lim_{x\to a^+}f'(x)
$$
is the limit of $f'(x)$ as $x$ goes to $a$ from the right (and similarly for $f'_-(a)$ and $f'(a-)$).
If $f$ is differentiable at $x=a$, then $f'_+(a)=f'_-(a)=f´(a)$. But if $f$ is diferentiable on a neighbourhood of $a$, $f'(a+)$ and $f'(a-)$ may not exist, as your example shows.
A: Oh dear, I believe I misread your question initially (I didn't see the primes somewhere).  If $f$ is differentiable at $x$, then $f'(x)$ need not be continuous at $x$.  However, of course, if $f'$ is continuous at $x$, then certainly $f$ will be differentiable at $x$.  After all, you need to know in the first place that $f'$ exists in a neighborhood of $x$ to say it is continuous there.  Your example demonstrates this perfectly.
ORIGINAL ANSWER TO MISREAD QUESTION:  If a function is differentiable at that point, then it must be continuous at that point.
Proof:  Suppose $\lim _{h\to 0}\frac{f(x+h)-f(x)}{h}$ converges.  Then,
$$
\lim _{h\to 0}\left[ f(x+h)-f(x)\right] =\lim _{h\to 0}\left[ h\left( \frac{f(x+h)-f(x)}{h}\right) \right] =0\cdot f'(x)=0.
$$
If a function is continuous at that point, it need not be differentiable. For example, $f(x)=|x|$ is continuous at $0$, but not differentiable.
