Awkward behavior of $x^2 \sin\frac{1}{x}$ at $x=0$? According to p. 107 of the book Advanced Calculus by Fitzpatrick, 

Assuming that the periodicity and differentiability properties of the sine function are familiar, the following is an example of a differentiable function having a positive derivative at $x = 0$ but such that there is no neighborhood of $0$ on which it is monotonically increasing:
  $$f(x) = \begin{cases}x^2\sin 1/x & \text{if}\ x\neq 0 \\
0 & \text{if}\ x = 0\end{cases}$$
  The source of this counterintuitive behavior is that the derivative $f'$ is not continuous at $x = 0$.

There are two confusing things about it: 


*

*a. By the definition of derivative (at $x=0$), $$\lim_{x\to 0} \dfrac{f(x)-f(0)}{x-0} = \lim_{x\to 0} \dfrac{x^2 \sin \bigl(\frac{1}{x}\bigr)}{x} = 0,$$ since $\bigl\lvert\sin\bigl(\frac{1}{x}\bigr)\bigr\rvert \le 1$. 
b. By use of rules for the derivative of products and quotients, $$f'(0) = \biggl[ 2x \sin \biggl(\frac{1}{x}\biggr) + x^2 \biggl(-\frac{1}{x^2} \cos \biggl(\frac{1}{x}\biggr)\biggr) \biggr]_{x=0} $$ which is not defined since $\cos\bigl(\frac{1}{0}\bigr)$ is not defined.
So why the text says that its derivative exists and its value is $>0$? 

*The function $f(x)$ is monotonically increasing because it is an odd function so for any neighborhood abound $x=0$, $f(x_1>0)>f(x_2<0)$. So why the text says otherwise? 

1.- (a) is correct so $f'(0)=0$, but your conclusion on 1.- (b) is wrong. Notice that you are trying to compute the derivative of $f$ at $0$ by using the formula for $f$ on points different from zero. The formula 
$$
f'(x) = 2x \sin \biggl(\frac{1}{x}\biggr) + x^2 \biggl(-\frac{1}{x^2} \cos \biggl(\frac{1}{x}\biggr)\biggr) 
$$
holds for all $x\neq0$ but, as you know, this formula is undefined at $0$. Therefore, you can conlude that the derivative of $f$ is not continuous at $0$, since you cannot approximate $f'(0)$ by $f'(x)$ for $x$ near $0$. In other words,
$$
f'(0) \neq \lim_{x\rightarrow 0} f'(x).
$$
2.-
In general, odd functions are not monotonically increasing. For example, 
$$
x^3 - x
$$
is odd, but it is not monotonic. Indeed, $x^3-x$ is increasing on the intervals $(-\infty,-\tfrac{1}{\sqrt{3}})$ and $(\tfrac{1}{\sqrt{3}},\infty)$, but is decreasing on $(-\tfrac{1}{\sqrt{3}},\tfrac{1}{\sqrt{3}})$.
Back to $f(x) = x^{2}\sin(\tfrac{1}{x})$, you can tell that it is not increasing because it is positive on the intervals $(\tfrac{1}{2k\pi},\tfrac{1}{(2k-1)\pi})$ but negative on $(\tfrac{1}{(2k+1)\pi},\tfrac{1}{2k\pi})$ for every $k\neq0$. Since every neighborhood of $0$ contains infinitely many intervals of those types, we conclude that $f(x)$ has infinitely many changes of sign on any neighborhood of $0$. Thus, it is not monotonic near $0$.
