For the functions $f(x)=\frac{x}{x+3}$ and $g(x)=\frac{13}{x}$, find the composition $f\circ g$ and simplify your answer as much as possible. Write the domain using interval notation.
$$(f-g)(x)=◻$$

Domain of $f=g:◻$
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Step 1 ：The composition of functions is a function that applies one function to the results of another. In this case, we need to substitute $g(x)$ into $f(x)$ to get $f(g(x))$.

Step 2 ：The domain of a function is the set of all possible input values (x-values) which will produce a valid output. To find the domain of the composition $f(g(x))$, we need to consider the domains of both $f(x)$ and $g(x)$, and exclude any x-values that would make either function undefined.

Step 3 ：For the function $f(x)=\frac{x}{x+3}$, the denominator cannot be zero, so $x\ne -3$. For the function $g(x)=\frac{13}{x}$, the denominator cannot be zero, so $x\ne 0$.

Step 4 ：When we substitute $g(x)$ into $f(x)$, we get $f(g(x))=\frac{13}{x\ast (3+13/x)}$.

Step 5 ：Simplifying the above expression, we get $f(g(x))=\frac{13}{3x+13}$.

Step 6 ：The denominator of the function $f(g(x))$ is zero when $x=-\frac{13}{3}$. So, the domain of $f(g(x))$ is all real numbers except $x=0$ (from the domain of $g(x)$) and $x=-\frac{13}{3}$ (from the domain of $f(g(x))$).

Step 7 ：In interval notation, this is $(-\mathrm{\infty },-\frac{13}{3})\cup (-\frac{13}{3},0)\cup (0,\mathrm{\infty })$.

Step 8 ：So, the composition $f(g(x))$ is $\frac{13}{3x+13}$ and its domain is $(-\mathrm{\infty },-\frac{13}{3})\cup (-\frac{13}{3},0)\cup (0,\mathrm{\infty })$.

Step 9 ：Final Answer: $(f-g)(x)=\boxed{{\displaystyle \frac{13}{3x+13}}}$ and Domain of $f=g:\boxed{{\displaystyle (-\mathrm{\infty },-\frac{13}{3})\cup (-\frac{13}{3},0)\cup (0,\mathrm{\infty })}}$.