I have had this question for quite some time now but never tried to look for the answer but now it bugs me.
When simplifying $y=\dfrac1{\frac1x}$ we get $y=x$ where $0$ is defined ,I presume it is because the denominator can not be zero in a fraction ? Even if that is true what is wrong with it being zero as we get zero eventually :
$$y=\frac{1}{\frac{1}{0}}$$
$$y=\frac{1}{\infty}=0$$
Sorry if this is very naive :(
On
Define the map $f$ on the non-zero reals as the reciprocal function $f(x) = \frac 1 x$, and let $g = f \circ f$, the composition of $f$ with itself. The domain and the target of $f$ are $\mathbb R \setminus \{0\}$ by definition, and thus the same must hold for $g$. Diagrammatically: $$\mathbb R\setminus\{0\} \xrightarrow{\ \ f\ \ }\mathbb R\setminus\{0\}\xrightarrow{\ \ f\ \ }\mathbb R\setminus\{0\}.$$ The observation you make is equivalent to noticing that $g$ acts as the identity on its domain; thus, it can be continuously extended to the whole of $\mathbb R$ by prescribing $g(0) = 0$.
On
The quotient $\frac10$ is undefined and $\infty$ is not a number. So, your justification is not appropriate.
It is usual, when we deal with rational fractions, to simplify them in following since: if you have, say, $\require{cancel}\frac{x^2-1}{x^2+x}$, you notice that this is equal to $\frac{\cancel{(x+1)}(x-1)}{\cancel{(x+1)}x}$, and then we simplify this, getting $\frac{x-1}x$. And this expression is defined for every number other than $0$.
In your case, after simplifying $\frac1{1/x}$ as much as possible, what we get is simply $x$. And $x$ is defined everywhere. So, we say that $\frac1{1/x}=x$ and leave it like that, without bothering with the number $0$.
I think we would have a removable discontinuity at $x=0$, and other than that the function would follow $y=x$. This is because, saying $\frac 10=\infty$ is not correct, as $\infty$ is not a number and hence cannot be equated to anything, in this sense. Rather, saying $\frac 10\to \infty$ makes sense. This itself points towards a removable discontinuity, as the limit at $x=0$ would exist, but not the functional value.