Answer
For $f+g, f-g, fg$: $D=\{x|x\in\mathbb{R}\}$,
For $\frac{f}{g}$: $D_{\frac{f}{g}}=\left\{x|x\in\mathbb{R}-\left\{ \pm \frac{1}{\sqrt 3}\right\}\right\}$
Work Step by Step
$f(x)=x^2+2x$,
$D_f=\{x|x\in\mathbb{R}\}$,
$g(x)=3x^2-1$,
$D_g=\{x|x\in\mathbb{R}\}$,
thus,
$f+g=x^2+2x+3x^2-1=4x^2+2x-1$,
$D_{f+g}=\{x|x\in\mathbb{R}\}$,
$f-g=x^2+2x-3x^2+1=-2x^2+2x+1$,
$D_{f-g}=\{x|x\in\mathbb{R}\}$,
$f \times g=(x^2+2x)(3x^2-1)=3x^4-x^2+6x^3-2x\\
=3x^4+6x^3-x^2-2x$
$D_{f \times g}=\{x|x\in\mathbb{R}\}$,
$\frac{f}{g}=\frac{x^2+2x}{3x^2-1}$,
$3x^2-1=0\Rightarrow x=\pm\dfrac{1}{\sqrt 3}$
$D_{\frac{f}{g}}=\left\{x|x\in\mathbb{R}-\left\{ \pm \frac{1}{\sqrt 3}\right\}\right\}$
