If $y=1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}+\dots \infty$ with $|x|>1$, then $\frac{dy}{dx}$ is:

The given series is an infinite Geometric Progression (GP) with first term $a=1$ and common ratio $r=1/x$.Since $|x|>1$, $|r|<1$, so the sum exists:
$$y=\frac{a}{1-r}=\frac{1}{1-1/x}=\frac{x}{x-1}$$
Now, differentiate $y$ with respect to $x$ using the quotient rule:
$$\frac{dy}{dx}=\frac{(x-1)(1)-x(1)}{(x-1{)}^2}=\frac{x-1-x}{(x-1{)}^2}=-\frac{1}{(x-1{)}^2}$$
From the expression of $y$:$y=\frac{x}{x-1}⟹\frac{y}{x}=\frac{1}{x-1}$ Squaring both sides:$\frac{y^2}{x^2}=\frac{1}{(x-1{)}^2}$ Therefore:$\frac{dy}{dx}=-\frac{y^2}{x^2}$