Let $$f(x) = \frac{x}{x-2}$$
Additionally, let $$g(x) = \frac{3}{x}$$
The domain of $f(x)$ is all reals, except for $x = 2$, and the domain for $g(x)$ is all reals, except for $x=0$.
$$(f \circ g)(x) = \frac{3}{3-2x}$$
However, the domain for this function seems to be not simply $x \ne 3/2$ but also $x \ne 0$, since it must comply with the domain of $g(x)$. However, the image of $(f\ o\ g)(x)$ does have a solution at $x = 0$, as one can also plainly tell from the function at face value. So how is this supposed to stay true? And why, in this case, need we only the domain restrictions for $g(x)$, and not say $f(x)$, saying that $x \ne 2$ as well?
$$(f\circ g)(x) = \frac{\frac{3}{x}}{\frac{3}{x}-2} = \frac{x}{\color{red}{x}}\frac{3}{3-2x}$$
So, $x$ must not be $0$ i.e. $0$ is not in the domain of $x$.