I think this can be shown using the chain rule, but I'm not confident that equality of functions implies equality of derivatives.
Assume continuity. Then we have
$D_xf(\frac{1}{x}) = f'(\frac{1}{x}) (-\frac{1}{x^2})$ using $f(\frac1x) = f(g(x))$ with $g(x) = \frac1x$ and then chain rule
Since $f'$ is discussed, then I assume that $f$ is differentiable on its domain. As a side note, its domain is not the set of real numbers. (Can you see why?) Thus, it is perhaps inappropriate to say "for all $x$," but I'll put that aside.
Now, putting $g(x)=\frac1x,$ we have that $f(x)=(f\circ g)(x)$ for all $x$ in the domain of $f.$ Applying Chain Rule (the hypotheses of which I leave to you to verify) does, indeed, give us $$f'(x)=f'\bigl(g(x)\bigr)\cdot g'(x)=f'\left(\frac1x\right)\cdot\frac{-1}{x^2}=-\frac1{x^2}f'\left(\frac1x\right).$$