Answer
$f$ is discontinuous at $x=2n\pi\pm\displaystyle\frac{3\pi}{2}$, where $n$ is integer.
Work Step by Step
\[y=f(x)=\frac{1}{1+\sin x}\]
Clearly $f$ is discontinuous at all values of $x$ for which denominator of $f$ is 0.
\[1+\sin x=0\Rightarrow \sin x=-1\Rightarrow \sin x=-\sin\frac{\pi}{2}\]
\[\Rightarrow \sin x=\sin\frac{3\pi}{2}\]
$\Rightarrow f$ is discontinuous at $x=2n\pi\pm\displaystyle\frac{3\pi}{2}$, where $n$ is integer.
Answer is : $f$ is discontinuous at $x=2n\pi\pm\displaystyle\frac{3\pi}{2}$, where $n$ is integer.