I am looking at the proof of the Stratonovich Chain rule in the book Klebaner, F.: Introduction to Stochastic Calculus with Applications, p. 146.
$\textbf{Theorem}$: Let $X$ be continuous and $f$ $\in$ $C^3$, then:
$f(X(t)) - f(X(0)) = \int_0^t f'(X(s)) \partial X(s) $,
where $\partial$ stands for the Stratonovich differential.
$\textbf{Proof}$: By Ito' s formula $f(X(t))$ is a semimartingale, and by definition of the stochastic integral $f(X(t)) - f(X(0)) = \int_0^t df(X(s))$. By Ito' s formula $$df(X(t)) = f'(X(t))dX(t) + \frac{1}{2}f''(X(t))d[X,X](t)$$ Therefore, from the definition of the Stratanovich integral for two adapted processes $X$ and $Y$:
$$ Y(t)\partial{X(t)} = Y(t)dX(t) + \frac{1}{2}d[Y,X](t)$$
only the equality $f''(X(t))d[X,X](t) = d[f'(X(t)),X](t)$ needs to be shown. By applying Ito' s formula on $df'(X(t))$ we obtain:
$$ df'(X(t)) = f''(X(t))dX(t) + \frac{1}{2}f'''(X(t))d[X,X](t)$$
and therefore by multiplying the terms from $d[f'(X(t)), X](t)$ with each other, we obtain:
$$d[f'(X(t)), X](t) = f''(X(t))dX(t)dX(t) = f''(X(t))d[X, X](t)$$
as needed.
$\textbf{Question}$ I don't understand why the multiplication of $dX(t)$ and $d[X, X](t)$ disappeared. I guess it's because the former term $(X(t))$ is continuous and the latter term is of finite variation $([X,X](t))$ (Theorem 1.11 in the same book specifies that covariation of continuous function $g$ and function $f$ of finite variation is zero) - however, I don't know of any theorem that would prove that for continuous functions the quadratic variation is of finite variation. Could you give me some advice on this part?