If $f'(a)$ exists, does $f'(a^+)$ and $f'(a^-)$ exist? Is it true that

If $f(x)$ is  differentiable at $a$, then both $f'(a^+)$ and $f'(a^-)$ exist and $f'(a^+)=f'(a^-)=f'(a)$.

My answer is NO.

Consider the function
$$
f(x)=\begin{cases}
x^2\sin\dfrac{1}{x}&\text{for $x\ne0$}\\[1ex]
0&\text{for $x=0$}
\end{cases}
$$
$f'(0)$ can be found by 
\begin{align} \lim_{x \to 0} \dfrac{f(x) - f(0)}{x-0} & = \lim_{x \to 0} \dfrac{f(x) - 0}{x}  & \textrm{ as } f(0) = 0 \\ & = \lim_{x \to 0} \dfrac{x^2 \sin\left(\frac{1}{x}\right)}{x} & \\
& = \lim_{x \to 0} x \sin\left(\frac{1}{x}\right) & \end{align}
Now we can use the Squeeze Theorem. As $-1 \leq \sin\left(\frac{1}{x}\right) \leq 1$, we have that $$0 = \lim_{x \to 0} x \cdot -1 \leq \lim_{x \to 0} x \sin\left(\frac{1}{x}\right) \leq \lim_{x \to 0} x \cdot 1 = 0$$
Therefore, $\lim_{x \to 0} x \sin\left(\frac{1}{x}\right) = 0$ and we have $f'(0)=0$. 

However, 
$$
f'(x)=\begin{cases}
-\cos\dfrac{1}{x}+2x\sin\dfrac{1}{x}&\text{for $x\ne0$}\\[1ex]
0&\text{for $x=0$}
\end{cases}
$$
$f'(0^+)$ nor $f'(0^-)$ exists as $x\to 0$.

Is my answer correct?

I have found some pages related to this question.
Is $f'$ continuous at $0$ if $f(x)=x^2\sin(1/x)$
Calculating derivative by definition vs not by definition
Differentiability of $f(x) = x^2 \sin{\frac{1}{x}}$ and $f'$
$f'$ exists, but $\lim \frac{f(x)-f(y)}{x-y}$ does not exist
Thanks.

Yes, your answer is correct. The existence of the derivative of a function at a point does not always mean that the derivative will be continuous at that point. The condition $f′(a+)=f′(a−)=f′(a)$  implies continuity of the derivative at $x=a$ which is clearly not true for the function you mentioned at $x=0$.
