Compute $f'(0)$ and check if $f'$ is continuous or not 
Given the function 
  $$f(x)=\begin{cases}x^{4/3}\sin(1/x)&\text{if $x\neq 0$}\\0&\text{if $x=0$}\end{cases}.$$
  
  
*
  
*Compute  $f'(0)$;
  
*Is $f'$ continuous on  $\mathbb{R}$.

I am unsure of how to solve this but will post what I have.
Computing the part 1 it shows that  $f'(0) =0$. (unless I computed it wrong)
For part two I am totally lost on how to prove this, but am thinking that since $f'=0$ wouldn't that make the function not continuous?. Any help is appreciated!
 A: Yes, $f'(0)=0$ is correct. On the other hand for $x\not=0$,
$$f'(x)=\frac{4}{3}x^{4/3-1}\sin(1/x)+x^{4/3}\cos(1/x)(-1/x^2).$$
It easy to verify that $f'$ is continuous in $\mathbb{R}\setminus\{0\}$.
So the answer to question 2 is yes if and only if we show that
$$\lim_{x\to 0}f'(x)=f'(0)=0.$$
A: $$f(x)=\begin{cases}x^{4/3}\sin(1/x)&\text{if $x\neq 0$}\\0&\text{if $x=0$}\end{cases}.$$
We have, 
$$f'(0)=\lim_{h\to 0}\frac{f(h)-f(0)}{h}=\lim_{x\to 0}h^{1/3}\sin\left(\frac{1}{h}\right) = 0$$
Since $|\sin\left(\frac{1}{h}\right)|\le 1$. Then 
$$f'(x)=\begin{cases}\frac43x^{1/3}\sin(1/x) - x^{-2/3}\cos(1/x)&\text{if $x\neq 0$}\\0&\text{if $x=0$}\end{cases}.$$
But $$\lim_{x\to 0}f'(x) =\lim_{x\to 0}\frac43x^{1/3}\sin(1/x) - x^{-2/3}\cos(1/x) \\= \color{red}{-\lim_{x\to 0} x^{-2/3}\cos(1/x)~~DNE}$$

does not exists . So $f'$ is not continuous at $x=0$.

Indeed let $g(x) =  x^{-2/3}\cos(1/x) $ and set $z_n = \frac{1}{n\pi}$  $$ \lim_{n\to \infty}z_n =0$$
But $$g(z_n) =  \left(n\pi\right)^{2/3} \cos\left(n\pi\right) = (-1)^n\left(n\pi\right)^{2/3} = \begin{cases} \left(2k\pi\right)^{2/3}& n=2k\\-\left((2k+1)\pi\right)^{2/3}& n=2k+1\end{cases}$$
we see that $$ \lim_{n\to \infty}g(z_n)~~~DNE$$
A: Similar problems like this one are around on the Internet.
For the first one you can do Squeeze theorem since the sine modulates between $-1$ and $1$. Second part: For the derivative you need to use the limit definition $\lim\limits_{x\to0}\frac{f(x)-f(0)}{x-0}$, using your answer from part a) for $f(0)$ and since the exponent is more than $1$, you will find that $f'(0)$ exists (although the derivative is NOT continuous) 
