Answer
$$(f+g)(x)=\frac{1+2x^2}{x+x^3}$$
$$(f-g)(x)=-\frac{1}{x+x^3}$$
$$(fg)(x)=\frac{1}{1+x^2}$$
$$(f/g)(x)=\frac{x^2}{1+x^2}$$
$$D_{f+g}=D_{f-g}=D_{fg}=\mathbb{R}- \{0 \}$$
$$D_{f/g}=\mathbb{R}- \{0 \}$$
Work Step by Step
$$(f+g)(x)=f(x)+g(x)=\frac{x}{1+x^2}+\frac{1}{x}=\frac{1+2x^2}{x+x^3}$$
$$(f-g)(x)=f(x)-g(x)=\frac{x}{1+x^2}-\frac{1}{x}=-\frac{1}{x+x^3}$$
$$(fg)(x)=f(x)g(x)=(\frac{x}{1+x^2})(\frac{1}{x})=\frac{1}{1+x^2}$$
$$(f/g)(x)=f(x)/g(x)=(\frac{x}{1+x^2})/(\frac{1}{x})=\frac{x^2}{1+x^2}$$
The domains of $f+g$, $f-g$, and $fg$ are the intersection of the domains of $f$ and $g$, and the domain of $f/g$ is the intersection of the domains of $f$ and $g$ minus those points vanishing $g(x)$. So we have
$$D_{f+g}=D_{f-g}=D_{fg}=\mathbb{R} \cap (\mathbb{R}- \{0 \} ) =\mathbb{R}- \{0 \}$$ $$ D_{f/g}=(\mathbb{R} \cap (\mathbb{R}- \{0 \} )-\varnothing =\mathbb{R}- \{0 \}.$$
