What is the Domain of $\frac{2}{5x+8}$ I had originally thought that it was $(-\infty, -8/5) \cup (-8/5,\infty)$ but LON CAPA says that is wrong. Original function is $$\frac{2(\frac{1}{x})}{8(\frac{1}{x})+5}$$
The question "What is the domain of [some expression]" is a poorly phrased question.  A function is defined, in part, by its domain.  If you are not told ahead of time what the domain (and codomain!) of a function is (are), then any answer is legit.  For example, the answer $\{1, 2, \pi, -10^{100}\}$ is a perfectly reasonable answer to the given question.  However, I am being cranky.  I presume that the question is meant to be read


  What is the largest set of real numbers on which the expression
  $$ \frac{2\cdot\frac{1}{x}}{8\cdot\frac{1}{x}+5} $$
  defines a function?

After doing a bit of simplification, you seem to have noticed that
$$ \frac{2\cdot\frac{1}{x}}{8\cdot\frac{1}{x}+5} = \frac{2}{5x+8}. $$
However, this equality holds only when both the left-hand side (LHS) and the right-hand side (LHS) are defined.  The two expressions are not defined for all of the same values (and, in fact, this is sufficient to show that the largest set of numbers on which the LHS defines a function is different from the largest set of real numbers on which the RHS defines a function).  The simplification step can be useful in other circumstances, but it isn't helpful for determining the maximal domain.
Finally, to get at the largest possible domain, note that the original expression is a fraction of fractions involving a variable $x$.  It will define a function as long as the fraction takes a well-defined value.  Typically, fractions take well-defined values whenever both their numerators and denominators are real numbers, and their denominators are nonzero.  Thus we have three questions to answer:


When is the numerator a real number?
When is the denominator a real number?
When is the denominator nonzero?


The numerator is a real number as long as $\frac{1}{x}$ is defined, which implies that $x\ne 0$.  Similarly, the denominator is defined as long as $x\ne 0$.  Thus we definitely know that $0$ is not in the domain of the function defined by this expression.  Finally, note that
$$ 8\cdot\frac{1}{x} + 5 = 0
\iff \frac{1}{x} = -\frac{5}{8}
\iff x = -\frac{8}{5}.
$$
Thus the fraction is undefined when $x = -\frac{8}{5}$, which implies that this value is not in the domain of the function.

We have shown that neither $0$ nor $-\frac{8}{5}$ is in the domain of the function, but every other real number works (since all other real numbers satisfy the conditions addressed by the three questions above).  Therefore the domain of the function is the set of real numbers, minus those give above, i.e.
$$ \mathbb{R} \setminus \left\{ 0, -\frac{8}{5} \right\} = \left( -\infty, -\frac{8}{5} \right) \cup \left(-\frac{8}{5}, 0\right) \cup (0,\infty).$$
$f(x)=\dfrac{2(\dfrac{1}{x})}{8(\dfrac{1}{x})+5}$

$\mathcal{D}_f = \bigg\{x\in \mathbb{R};\quad x\ne0\land \left(\dfrac{8}{x}+5\right)\ne 0\bigg\}$