I'm a bit embarrassed to admit that:

a) This is a rather unmotivated question.

b) I can't remember whether or not I've asked this before, but searching doesn't seem to turn anything up so ...

Consider some "shape" function $\phi: \mathbf{R} \to \mathbf{R}$. Then given some function $f: \mathbf{R} \to \mathbf{R}$, one can ask whether the "difference quotient",

$\lim_{y\to x} \frac{f(y)-f(x)}{\phi(y-x)}$,

exists at various points $x$. Letting $\phi(x) = x$ corresponds to taking normal derivatives, and intuitively when the limit exists this means that near $x$, the function $f$ "looks like" $\phi$ does near 0.

However, if the ratio $\phi(x)/x$ is not bounded above or away from 0 as $x\to 0$ (I'm mostly thinking of the case when it is neither, so that $\phi$ is "wildly oscillating" in some sense), then anywhere the above limit exists and is nonzero, the function $f$ is necessarily non-differentiable.

My question: If $\phi$ is some wildly oscillating function as described above (pick your favorite), can there be an $f$ for which this limit exists everywhere?

(Edit: I suppose I really want $\phi$ and $f$ to be continuous functions.)