Ok Let talk about this,... I am now so confused.
1-$$\mathcal{F}\Big\{c(x-x_0)b(x-x_0)\Big\}=\mathcal{F}\Big\{c(x-x_0)\Big\}\circ\mathcal{F}\Big\{b(x-x_0)\Big\}\\=\Bigg[e^{-2ix_0y}C(y) \Bigg]\circ\Bigg[e^{-2ix_0y}B(y)\Bigg]\\ =\Bigg(e^{-2ix_0y}\Bigg)^2\Bigg[C(y) \circ B(y)\Bigg]$$ or
2-$$\mathcal{F}\Big\{c(x-x_0)b(x-x_0)\Big\}=\mathcal{F}\Big\{(cb)(x-x_0)\Big\}\\= e^{-2ix_0y}\Bigg(\mathcal{F}\Big\{(cb)(x)\Big\}\Bigg)\\ =e^{-2ix_0y}\Bigg[C(y) \circ B(y)\Bigg]$$
Please which of these two is valid? and if is (2) why the first is not exact? What error I am making? Thank you so much
Note that $\mathcal{F}$ is the Fourier transform and $\circ$ is convolution.
Let $f(x) = c(x) \cdot b(x)$, then $f(x-x_0) = f(x) \circ \delta(x - x_0) = (c(x) \cdot b(x) ) \circ \delta(x - x_0)$. The Fourier transform is then
$$ \begin{align} \mathcal{F}\{f(x)\} & = \mathcal{F}\{ f(x) \circ \delta(x - x_0) \} \\ & = \mathcal{F}\{f(x)\} \cdot \mathcal{F}\{\delta(x - x_0)\} \\ & = \mathcal{F}\{c(x) \cdot b(x)\} \cdot e^{-j2\pi y x_0} \\ & = \left( \mathcal{F}\{c(x)\} \circ \mathcal{F}\{b(x)\} \right) \cdot e^{-j2\pi y x_0} \\ & = \left( C(y) \circ B(y) \right) \cdot e^{-j2\pi y x_0}. \end{align} $$
I believe your mistake in the first one is that you are treating $e^{-j2\pi y x_0}$ as a scalar in order to bring it out of the convolution. However, it is not a scalar. Rather, it's a function of the frequency $y$. That is, I believe the following is the case:
$$ \left( f(t) \cdot g(t) \right) \circ \left( f(t) \cdot h(t) \right) \neq f(t) \cdot \left( g(t) \circ h(t) \right) $$