We have $$\sqrt{d} = \frac{x}{y} - \frac{1}{f_0\cdot y} - \frac{1}{f_0\cdot f_1\cdot y}- \ldots - \frac{1}{f_0\cdot f_1\cdot\ldots\cdot f_n\cdot y}-\ldots\,,$$ where $$f_0 = 2x\,,$$ $$f_{n+1} = (f_n)^2 - 2\text{ for }n=0,1,2,\ldots,$$ and $x$ and $y$ are nontrivial solutions to Pell's Equation $x^2 - d\cdot y^2 = 1$.
For example:
$$\sqrt{5} = 9/4 - 1/72 - 1/23184 - 1/2043763488 -\ldots\,.$$
I don't know if it is published somewhere, but in my opinion it has little to do with Pell equation.
From $x^2-d y^2=1$ you deduce $d=\left(\frac xy\right)^2-\frac1{y^2}=\left(\frac xy\right)^2\left(1-\frac1{x^2}\right)$, hence $\sqrt d = \frac xy \sqrt{1-\frac1{x^2}}$. Your identity can be written as $$ \sqrt{d} = \frac xy \left(1-\frac1{f_0x}-\frac1{f_1f_0x}-\frac1{f_2f_1f_0x}-\dots\right), $$ therefore it is basically stating that $$ \sqrt{1-\frac1{x^2}} = 1-\frac1{f_0x}-\frac1{f_1f_0x}-\frac1{f_2f_1f_0x}-\dots $$
This is just an asymptotic expansion of the square root, and is completely decoupled from the original Pell equation.