Suppose $y$, Does $\frac{dy}{dy} $ have meaning when we derive it with respect to itself?

We are basically asking what is the rate by which $y$ changes with respect to $y$? Since $y$ changes proportionately to itself, the value is $1$.

Notice that you probably do this implicitly if you have differentiated some functions before. For instance, if $y=2x$, then $\displaystyle\frac{d}{dx}(y)=\frac{d}{dx}(2x)=2.$ That is, the rate by which $y$ changes with respect to $x$ is $2$, since every time $y$ increases or decreases by $2$, $x$ only increases or decreases by $1$. Now suppose instead that $y=x$, since $y$ is changing only as much as $x$ is, the rate by which $y$ changes with respect to $x$ is now $1$. Thus, $\displaystyle\frac{d}{dx}(y)=\frac{d}{dx}(x)=1$.

Define the function $f(y)=y$ $\;\;\;\;\;\;$ (1)

We will use the first principle here,

Let a small increment in $y$ correspond to $f(y+\Delta y)$

Then , $f(y+\Delta y)=y+\Delta y$ $\;\;\;\;\;\;\;$ (2)

Subtract (1) from (2)

$f(y+\Delta y)-f(y)=\Delta y$

Re-arranging.

$\frac{f(y+\Delta y)-f(y)}{\Delta y}=1$

Taking limit $\Delta y\to 0$

$\frac{d[f(y)]}{dy}=1=\frac{d(y)}{dy}$

Hence, essentially you are measuring the rate of change of quantity with respect to itself. It's quite easy to see why it makes sense.

Yes , example y=y, dy/dy=1 , means that y is equal to its own coordinate and its rate of change is 1