Consider two integrable functions $f(x)$ and $g(x)$, whose derivatives (and higher order derivatives) are also integrable. Note that $f, g: \mathbb{R} \to \mathbb{R}$. From the properties of the Fourier transform (and convolutions), the Fourier transform of the product $\frac{df}{dx}\, g(x)$ is
$$ \begin{align*} \mathcal{F}\left[\frac{df}{dx}\, g(x)\right](\omega) &= \mathcal{F}\left[\frac{df}{dx}\right](\omega) * \mathcal{F}\left[g(x)\right](\omega) \\ &= \left(2\,\pi\,i\,\omega\,\mathcal{F}\left[f(x)\right](\omega)\right) * \mathcal{F}\left[g(x)\right](\omega) \\ &= 2\,\pi\,i\,\omega\,\left(\mathcal{F}\left[f(x)\right](\omega) * \mathcal{F}\left[g(x)\right](\omega)\right) \\ &= 2\,\pi\,i\,\omega\,\left(\mathcal{F}\left[g(x)\right](\omega) * \mathcal{F}\left[f(x)\right](\omega)\right) \\ &= \left(2\,\pi\,i\,\omega\,\mathcal{F}\left[g(x)\right](\omega)\right) * \mathcal{F}\left[f(x)\right](\omega) \\ &= \mathcal{F}\left[\frac{dg}{dx}\right](\omega)* \mathcal{F}\left[f(x)\right](\omega) \\ &= \mathcal{F}\left[f(x)\right](\omega) * \mathcal{F}\left[\frac{dg}{dx}\right](\omega) \\ &= \mathcal{F}\left[f(x)\,\frac{dg}{dx}\right](\omega), \end{align*} $$
where $*$ specifies convolution.
Taking the inverse transform of both sides, we see that $\frac{df}{dx}\, g(x) = f(x)\,\frac{dg}{dx}$ which is obviously wrong. The issue is that I don't see what's wrong with the argument? The only thing I can think of is the integrability of $f(x)$ and $g(x)$ and their derivatives.
For context, I have been working with finite difference and pseudo-spectral methods, and wanted to investigate the Fourier transform of the Jacobian operator $J(f(x, y), g(x, y)) = \frac{\partial f}{\partial x}\,\frac{\partial g}{\partial y} - \frac{\partial f}{\partial y}\,\frac{\partial g}{\partial x}$.