Why am I getting derivative of $y = 1/x$ function as $0$?

The view of standard analysis has been clarified well.

Here is a explanation in the view of non-standard analysis. If $aapprox b$, is $frac{1}{a}approx frac{1}{b}$? $aapprox b$ means that $|a-b|$ is small enough. After multiplying $|frac{1}{ab}|$, we get $|frac{1}{a}-frac{1}{b}|$. But one cannot assure that this is also sufficiently small, this is because $|frac{1}{ab}|$ can be infinitely large.

To be precise, small enough means smaller than any $frac{1}{n}$, where $n$ is a positive integer, and large enough means that larger than any positive integers. The only small enough (or, say infinitesimal) element in the real number is nothing but $0$, but there are more in the hyperreal number.

Why it doesn't work can be seen by quantifying how accurate the approximation $f(x+y) approx f(x)$ is. This is rightfully a zeroth order Taylor expansion, $$ f(x+dx) = f(x) + O(dx) $$ for the function $f(x)=1/x$, we see $$ frac1{x+dx}-frac1x = O(dx)$$ and therefore begin{align*} frac{frac{1}{x+dx} - frac{1}{x}}{dx} &=left(frac{1}{x+dx} - frac{1}{x} right) frac{1}{dx} \ &= O(dx)frac1{dx} \&= O(1). end{align*} $O(1)$ quantities are not zero as $dxto0$, so we cannot conclude as you did. Instead, we need to use a more accurate expansion like a first order Taylor expansion (but that's a little circular), or be more careful in your algebraic manipulations like the other answers.

$$frac{,d}{,dx} frac{1}{x}=lim_{hrightarrow 0} frac{frac{1}{x+h}-frac{1}{x}}{h}= lim_{hrightarrow 0}frac{x-(x+h)}{hx(x+h)}=lim_{hrightarrow 0} -frac{1}{x(x+h)}=-frac{1}{x^2}$$

Do not remove $(dx)^2$: $$frac{frac{1}{x+dx} - frac{1}{x}}{dx}=frac{x-(x+dx)}{(x+dx)xdx}=-frac{1}{x(x+dx)} to -frac{1}{x^2}$$