Answer
$[-3,1)\cup(1+\infty)$
Work Step by Step
Interpret $f(x) $ as $\displaystyle \frac{u(x)}{v(x)}.$
The domain of f is the intersection of domains of u and v, with $v(x)\neq 0.$
Domain of $u(x)=\sqrt{x+3}$ is $x+3\geq 0$,
or, the interval $[-3,+\infty)$
The domain of $v(x)$ is all reals, from which we exclude $x=1$, as it yields zero in the denominator.
The domain of f is then, the interval $[-3,+\infty)$ excluding $x=1,$ that is,
$[-3,1)\cup(1+\infty)$