Let $f,g \in C(\mathbb{R})$ such that for all $x\in Z$ where $Z$ is some arbitrary set, $f(x)=g(X)=0$.
Is$$ h(x):=\begin{cases} \frac{f(x)}{g(x)}&, \; x\notin Z\\ 0 &, \; x\in Z \end{cases}$$ continuous? If not, is there some restriction on the set $Z$ or on the functions $f$ and $g$ so that $h(x)$ is then continuous?
Obviously, there are restrictions that can be placed on $f$ and $g$ that will guarantee that $h$ is continuous. For example, restrict $f(x) = x^2, g(x) = x$, and hey, $h(x) = x$, which is continuous! This is why Jacky Chong calls the problem too broad.
But it is likely that the question is looking for the loosest possible condition that can be placed on the functions and still make $h$ continuous. For that, we look at what is needed for $h$ to be continuous. Obviously for $x \notin Z, h$ is continuous at $x$. When $x \in Z$, we know that $h(x) = 0$, so continuity requires $\lim_{t \to x} h(t) = 0$. If it happens that $x$ is in the interior of $Z$ (i.e., there is some interval about $x$ which lies entirely within $Z$), then this limit is also not a problem as $h(t) = 0$ when $t$ gets close enough to $x$.
So the problem occurs when $x$ lies on the boundary of $Z$ (i.e. for any given distance from $x$, there are points outside of $Z$ that are at least that close to $x$). In this case, we have to require $$\lim_{t\to x, t\notin Z} \frac{f(t)}{g(t)} = 0$$ I.e., $f$ has to go to $0$ faster than $g$ does.