Here's an intuitive way of thinking about the problem.
(1) The $x^2$ on the outside causes the function to vanish rapidly, but the $1/x^2$ inside the sine function causes the oscillation to be similarly rapid. This balance turns out to be just enough to produce unbounded variation, as the variation behaves similarly to the harmonic series. How so?
It will suffice to consider the interval $[0,1]$. $\sin(1/{x}^2)= 0$ when
\begin{equation}x = \frac{1}{\sqrt{\pi n}}\end{equation}
and $\sin(1/x^2) = 1$ when
\begin{equation}x = \frac{\sqrt{2/\pi}}{\sqrt{4n+1}}\end{equation}
(in both cases, make sure that the denominators are not zero)
Now, "throw" these points into a partition. In other words, create a sequence of partitions containing these values for greater and greater $n$. Now, compute the variation. For the points of the form
\begin{equation}x = \frac{1}{\sqrt{\pi n}}\end{equation}
the entire expression vanishes. Thus the variation just becomes the summation of $x^2$ at points of the form
\begin{equation}x = \frac{\sqrt{2/\pi}}{\sqrt{4n+1}}\end{equation}
which is
\begin{equation}x = \sum_{n=n_{0}}^{k} \frac{2/\pi}{4n+1}\end{equation}
which, like the harmonic series, diverges as $k \to \infty$.
(2) In this case, the function vanishes at a speed faster than which it oscillates. This will give us bounded variation, in the form similar to that of the convergent sum $\sum 1/n^2$.
It will suffice to consider the interval $[0,1]$, as the mirror case is identical.
$\sin(1/x)= 0$ when
\begin{equation}x = \frac{1}{\pi n}\end{equation}
and $\sin(1/x) = 1$ when
\begin{equation}x = \frac{2}{\pi (4n+1)}\end{equation}
again making sure that the denominator is nonzero. Using the same technique as before, we construct a sequence of partitions where the $\sin(1/x)$ term either vanishes or equals one. The variation (of a particular partition in the sequence) is then the following sum
\begin{equation}x = \sum_{n=n_{0}}^{k} \frac{4}{\pi^2 (4n+1)^2}\end{equation}
which converges as $k \to \infty$ like $\sum 1/n^2$.
This technique can be extended to the "general" case of $x^{k}\sin(1/x^{n})$ very easily, and provides an interesting parallel between the vanishing/oscillation speeds of this function, and summations of the form $\sum 1/x^{m}$.
Take $a=0$, $b=1$,
$$f(x) := \begin{cases} x^2 \cdot \sin x^{-\frac{3}{2}} & x \in (0,1] \\ 0 & x=0 \end{cases}$$
Then $f$ is differentiable and of bounded variation, but $f'$ is unbounded.
Hint To show that $f$ is of bounded variation you can use the following theorem: Let $f: [0,1] \to \mathbb{R}$ differentiable and $f' \in L^1([0,1])$. Then $f$ is of bounded variation and $$\text{Var} \, f = \int_0^1 |f'(t)| \, dt$$
Remark As Pavel M suggested one can also prove that $f$ is of bounded variation by splitting up the interval $[0,1]$ in intervals $[a_n,b_n]$ such that $f$ is monotone on $[a_n,b_n]$. Then one can easily compute the variation of $f$ on the interval $[a_n,b_n]$ and use the fact that the variation on $[0,1]$ is equal to the sum of the variations on $[a_n,b_n]$.
Best Answer
$f(x)=\sin\left(x\sin x\right)$ is bounded and differentiable everywhere on $\mathbb{R}$, but it has not a bounded derivative.
Edit: To prove that $f'(x)=\cos(x\sin x)[\sin x+x\cos x]$ is unbounded, observe that $x\sin x=0$ when $x=2n\pi$ and $x\sin x=(2n+\frac12)\pi$ when $x=(2n+\frac12)\pi$. Hence there exists a sequence of positive numbers $x_n\to\infty$ such that $x_n\sin x_n=1$. Consequently, $$ \frac{f'(x_n)}{\cos 1} =\frac1{x_n}+x_n\sqrt{1-\frac1{x_n^2}}\to\infty \ \text{ as }\ n\to\infty. $$