Hyper Derivative definition.

What is $\boldsymbol{\,f^\nabla(x)}$? $$ \begin{align} f^\nabla(x) &=\lim_{h\to1}\log_h\left(\frac{f(xh)}{f(x)}\right)\\ &=\lim_{h\to1}\frac{\log(f(xh))-\log(f(x))}{\log(h)}\\ &=\lim_{\delta\to0}\frac{\log(f(x+\delta\,x))-\log(f(x))}{\log(1+\delta)}\\ &=\lim_{\delta\to0}\frac{\delta\,x}{\log(1+\delta)}\,\lim_{\delta\to0}\frac{\log(f(x+\delta\,x))-\log(f(x))}{\delta \,x}\\ &=x\,\frac{\mathrm{d}}{\mathrm{d}x}\log(f(x))\\ &=x\,\frac{f'(x)}{f(x)} \end{align} $$ which is well-defined as long as $\,f'(x)$ exists and $f(x)\ne0$.

Furthermore, if $f'(0)$ exists and $f(0)=0$, then $f^\nabla(0)=1$.

Geometric Interpretation

Extending the result above, $$ \begin{align} f^\nabla(x) &=x\,\frac{\mathrm{d}}{\mathrm{d}x}\log(f(x))\\ &=\frac{\mathrm{d}\log(f(x))}{\mathrm{d}\log(x)} \end{align} $$ which would be the slope of the graph of $f(x)$ in a log-log plot.

Assuming that $f$ is differentiable in the usual sense in $x$, $$ f(xh)=f(x)+(h-1)xf'(x)+o(h-1). $$ If $f(x)\neq 0$, then $ f(xh)/f(x)=1+(h-1)xf'(x)/f(x)+o(h-1)$. Moreover, $\ln h=h-1+o(h-1) $, hence $$ \log_h\frac{f(xh)}{f(x)}=\ln\frac{f(xh)}{f(x)}\cdot (\ln\,h)^{-1}=\frac{(h-1)xf'(x)/f(x)+o(h-1)}{h-1+o(h-1) }=\\ =x\frac{f'(x)}{f(x)}+o(h-1). $$ Therefore $$ f^\nabla(x)=x\cdot (\ln f)'(x). $$ This yields easily $(f\cdot g)^\nabla = f^\nabla + g^\nabla$ and provides the geometrical meaning in terms of the one of the derivative.

Let's say $f$ is continuous, $x>0$ and $f(x)>0$. Then, you are basically calculating $$\lim_{h\to 1} \frac{\ln f(xh)-\ln f(x)}{\ln h}=\lim_{u:=\ln h\to 0}\frac{\ln f(x\cdot e^u)-\ln f(x)}{u}=\lim_{u\to 0}\frac{\ln f(e^{u+\ln x})-\ln f(e^{\ln x})}{u}=\frac{d(\ln\circ f\circ\exp)}{dy}(\ln x)$$

So, for $x>0$ and $f(x)\ne 0$, you have that $f^\nabla(x)=g'(\ln x)$, where $g(y)=\ln \lvert f(e^y)\rvert$

Similarly, for $x<0$ and $f(x)\ne0$, then $f^\nabla (x)=g'(\ln x)$, where $g(y)=\ln\lvert f(-e^y)\rvert$.

If $f(x)=0$, then the $f^\nabla$ is not defined. And, clearly, if $f(0)\ne 0$, then $f^\nabla(0)=0$.