It is known that a function with bounded variation is differentiable almost everywhere.
There are also functions with unbounded variation that are differentiable almost everywhere (e.g. take $f:[0,1]\rightarrow\mathbb{R}$ with $f(0)=0$ and $f(x)=1/x$ otherwise).
But does there exist a function on $[0,1]$ with unbounded variation that is everywhere differentiable?
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Sticking with the example $f(x) = x^2\sin (1/x^2),$ which is everywhere differentiable, I think it's easier to deal with its total variation directly from the definition. Let $a_n = 1/\sqrt {\pi/2+2n\pi},\, b_n = 1/\sqrt {2n\pi}.$ Then
$$\sum_{n=1}^{\infty}|f(b_n)-f(a_n)| = \sum_{n=1}^{\infty}|f(a_n)|= \sum_{n=1}^{\infty}\frac{1}{\pi/2+2n\pi}.$$
Apply the limit comparsion test, using the terms $1/n$ for comparison, to see this series diverges. Since the total variation of $f$ on $[0,1]$ is at least the $\sup$ of all partial sums of this infinite series, we see $f$ has unbounded total variation on $[0,1].$
Consider $f:[0,1]\to\mathbb{R}$ by $f(x)=x^{2}\sin(\frac{1}{x^{2}})$ for $x\neq0$ and $f(0)=0$.
$\int_{0}^{1}\lvert f'(x)\rvert dx=\int_{0}^{1}\lvert2x\sin(\frac{1}{x^{2}})-\frac{2}{x}\cos(\frac{1}{x^{2}})\rvert\ge\int_{0}^{1}\frac{2}{x}\lvert\cos(\frac{1}{x^{2}})\rvert dx-\int_{0}^{1}2x\vert\sin(\frac{1}{x^{2}})\rvert$
$\ge\int_{0}^{1}\frac{2}{x}\lvert\cos(\frac{1}{x^{2}})\rvert dx-1$.