Answer
$f'(0)$ exists and $f'(0) = 0$
Work Step by Step
$f(x) = x^2~sin(\frac{1}{x})~~~$ if $x\neq 0$
$f(x) = 0~~~~~~~$ if $x = 0$
We can try to evaluate $f'(0)$ using Definition 4:
$\lim\limits_{h \to 0} \frac{f(0+h)-f(0)}{h}$
$=\lim\limits_{h \to 0} \frac{(0+h)^2~sin(\frac{1}{0+h})-0}{h}$
$=\lim\limits_{h \to 0} \frac{h^2~sin(\frac{1}{h})}{h}$
$=\lim\limits_{h \to 0} h~sin(\frac{1}{h})$
Note that for all $h$ such that $h \neq 0$:
$h(-1) \leq h~sin(\frac{1}{h}) \leq h(1)$
$-h \leq h~sin(\frac{1}{h}) \leq h$
Also:
$\lim\limits_{h \to 0} (-h) = \lim\limits_{h \to 0} h = 0$
According to the Squeeze Theorem:
$\lim\limits_{h \to 0} h~sin(\frac{1}{h}) = 0$
$f'(0)$ exists and $f'(0) = 0$
