Answer
$f^{\prime}(0)$ does not exist.
Work Step by Step
$\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x-0}=\lim _{x \rightarrow 0} \frac{x \sin \frac{1}{x}-0}{x-0}=\lim _{x \rightarrow 0} \frac{x \sin \frac{1}{x}}{x}=\lim _{x \rightarrow 0} \sin \frac{1}{x}$
We are going to show that $\lim _{x \rightarrow 0} \sin \frac{1}{x}$ does not exist. Consider the function $\sin \frac{1}{x}=g(x) .$ Take an arbitrary $L \in \mathbb{R}$
If $L \neq 1 .$ Take $\epsilon=\frac{|1-L|}{2}$
since $\lim _{y \rightarrow+\infty} \frac{2}{(2 y-1) \pi}=0$ we have that for every $\delta>0$ there exists $A>0$ such that $0A$
Because of what we did above, for each $\delta>0$ we can take an odd number $n_{\delta}>0$ such that $00$ take $x_{\delta}=\frac{2}{\left(2 n_{s}-1\right) \pi}$. We have that $g\left(x_{6}\right)=\sin \frac{1}{x_{6}}=$
$\sin \left(\frac{1}{\frac{2}{\left(2 n_{\delta}-1\right) \pi}}\right)=\sin \left(\frac{\left(2 n_{\delta}-1\right) \pi}{2}\right)=\sin \left(n_{\delta} \pi-\frac{\pi}{2}\right)=1$
So for each $\delta>0$ there exists $x_{6}>0$ such that $0\epsilon$ which gives us that $\lim _{x \rightarrow 0} \sin \frac{1}{x} \neq L$
If $L=1$ take $\epsilon=1$ , do the same as above but instead of taking odd $n_{\delta}^{\prime} s$ take them even. Hence you're gonna have $g\left(x_{8}\right)=-1$ and $\left|g\left(x_{8}\right)-L\right|=$ $|-1-1|=2>\epsilon$, then $\lim _{x \rightarrow 0} \sin \frac{1}{x} \neq 1$
Therefore you have that no $L \in \mathbb{R}$ can be $\lim _{x \rightarrow 0} \sin \frac{1}{x}$ which gives us that $f^{\prime}(0)$
does not exist.
