I tried evaluating the integral:
$$ \hat{g}(w) = \int_{-\infty}^{+\infty} g(x) e^{-iwx} dx = \int_{-\infty}^{+\infty}f(\alpha x) e^{-iwx}dx = \frac{1}{\alpha}\int_{-\infty}^{+\infty}f(u)e^{-\frac{iwu}{\alpha}} du $$
I don't know what to do about the $\frac{1}{\alpha}$ in the exponential, or where the $|\alpha|$ in the final result is supposed to come from.
If $\alpha >0$ then what you have done is correct and the integral you have obtained is simply the definition of the Fourier transform of $f$ evaluated at the point $\frac w {\alpha}$. If $\alpha <0$ then the integral on the right side would be from $\infty$ to $-\infty$, not $-\infty$ to $\infty$. That is where you get a minus sign and $\alpha$ in the denominator becomes $-\alpha$ or $|\alpha|$. For $\alpha =0$ $g$ has no Fourier transform unless $f=g=0$.