Let us say that a function is "discontinuous at $a$" if it is defined but not continuous at $a$.
Let $f$ and $g$ be real valued functions of real variable, and let $a$ be a real number. Here are a few things that are true:
If $f$ and $g$ are continuous at $a$, then so are $\alpha f$ (for any real number $\alpha$), $f+g$, $f-g$, $fg$, and if $g(a)\neq 0$ also $\frac{f}{g}$.
If $f$ is continuous at $a$ and $g$ is discontinuous at $a$, then $f+g$, $f-g$, and $\alpha g$ (for any real number $\alpha\neq 0$) are discontinuous at $a$.
If $f$ is continuous at $a$, $g$ is discontinuous at $a$, and $f(a)\neq 0$, then $fg$ is discontinuous at $a$.
If $g$ is discontinuous at $a$, and there is an open interval containing $a$ where $g$ is never equal to $0$, then $\frac{1}{g}$ is discontinuous at $a$.
For example, to show that if $f$ is continuous at $a$ and $g$ is discontinuous at $a$, then so is $f+g$, note that if $f+g$ were continuous at $a$, then $(f+g)-f$ would also be continuous at $a$. But $f+g-f = g$, which is discontinuous at $a$.