Are sets with 0 as an element not allowed to be a domain for f(x) = 1/x?

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Does a domain for a function mean that all its members have a defined output value? If this is so, is the distinction between a total function vs a partial function relevant to some arbitrary set, not to the domain itself?

Also, if a function is a total function of some set as its domain, does that mean that the set is a computable set? Is it impossible for a total function to not have an algorithm?

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Does a domain for a function mean that all its members have a defined output value?

Yes, pretty much by definition. So the answer to question in the title is yes, as $1/0$ is not defined.

If this is so, is the distinction between a total function vs a partial function relevant to some arbitrary set, not to the domain itself?

The writing $f:\subset X \to Y$ (i.e. $f$ is a partial function from $X$ to $Y$) means that $\mathrm{dom}(f)\subset X$. Every partial function $f$ is a total function $f:\mathrm{dom}(f)\to Y$.

Also, if a function is a total function of some set as its domain, does that mean that the set is a computable set? Is it impossible for a total function to not have an algorithm?

Computability has nothing to do with totality. You can have total non-computable functions and partial computable functions over a non-computable set. E.g. the characteristic function $\chi_A$ of a non-computable set $A$ is total but not computable. On the other hand, the restriction of $\chi_A$ to $A$ is constantly $1$, hence computable.