Why is $\frac{1}{\frac{1}{X}}=X$?

$1/x$ is, by definition, the number that you multiply $x$ by to get $1$.

Similarly, $1/\left(1/x\right)$ is the number that you multiply $1/x$ by to get $1$.

But wait a sec: we just learned in the first sentence that that number is $x$.

Maybe this will help you see why $\;\dfrac 1{\large \frac 1X}= X.\;$ We multiply numerator and denominator by $X$, which we can do because we can multiply any number by $\dfrac XX = 1$ without changing the actual value of the number:

$$\frac 1{\Large \frac 1X}\cdot \frac XX = \frac X1 = X$$ $$ $$

$$y=\frac1{\frac 1 x} $$ $$y'(x)=\left(\frac1{\frac 1 x}\right)' = -\left(\frac 1 {\left(\frac 1x\right)^2}\right)\left({\frac 1 x}\right)' = \frac 1 {\left(\frac 1x\right)^2} \cdot {\frac 1 {x^2}} = \frac {y^2(x)}{x^2}$$

So we have that $$x^2dy = y^2dx\\ \int \frac{dy}{y^2} = \int \frac{dx}{x^2}\\ -\frac{1}{y} = -\frac 1 x + C\\$$

Let's take a look at $y(1)$. $\frac 1 1 = 1$, this is already explained in a more common problem here: Why is $n$ divided by $n$ equal to $1$? So $y(1)=\frac{1}{\frac{1}{1}} =\frac 1 1 = 1$.

Note that I lost one possible solution, $y(x)=0$, by dividing by $y$. But since $y(1)=1$, it isn't really the solution.

Again: $y(1)=1$, so $~-\frac 1 1 = -\frac 1 1 + C ~~\Rightarrow~~ C=0$. Then $\frac {1} {y} = \frac {1}{x} \Rightarrow x=y$.