For $f(x)$ we have that
$$ \widehat{f}(\xi) = \int_\mathbb{R} e^{-2 \pi i x \xi} f(x) $$
Now let $s \in \mathbb{R}$ and consider
$$ \widehat{f(sx)}(\xi) = \int_\mathbb{R} e^{-2 \pi i (sx) \xi} f(sx) = \int_\mathbb{R} e^{-2 \pi i (x) \xi} f(x)*{d \over ds}(1/s) = -{1 \over s^2}\int_\mathbb{R} e^{-2 \pi i (x) \xi} f(x) = -{1 \over s^2} \widehat{f}(\xi) $$
through a change in variable $x \mapsto (1/s)x$.
Question: Is this calculation correct?
No. Let $\varphi(x)=sx$, so the question is addressing the relationship between $\widehat{f}$ and $\widehat{f\circ\varphi}$. We have \begin{align*} \widehat{f\circ\varphi}(\xi)&=\int f\circ\varphi(x)e^{-2\pi ix\xi}dx\\ &=\int f(sx)e^{-2\pi ix\xi}dx\\ &=\int f(sx)e^{-2\pi i(sx)(\xi/s)}\dfrac{d(sx)}{s}\\ &=\dfrac{\text{sgn}(s)}{s}\int f(x)e^{-2\pi ix\xi/s}dx\\ &=\dfrac{\text{sgn}(s)}{s}\widehat{f}(\xi/s). \end{align*}