I am investigating the following coupled sequence: \begin{align*} y_0 &= 1\\ x_{n+1} &= \sqrt{1 + \frac{1}{y_n}}\\ y_{n+1} &= \sqrt{1 - \frac{1}{x_{n+1}}}\\ \end{align*} I am trying to show \begin{align*} \lim_{n \to \infty} \frac{\left |x_{n+1} - \varphi\right |}{\left |x_n - \varphi\right |} &= \frac{1}{4}\\ \lim_{n \to \infty} \frac{\left |y_{n+1} - \varphi^{-1}\right |}{\left |y_n - \varphi^{-1}\right |} &= \frac{1}{4} \end{align*} where $\varphi = \frac{-1 + \sqrt{5}}{2}$.
So far, I've shown: \begin{align*} x_{n+1}^2 - \varphi^2 &= 1 + \frac{1}{y_n} - \varphi^2\\ &= \sqrt{1 + \frac{1}{x_n - 1}} - \varphi\\ &= \frac{1 + \frac{1}{x_n - 1} - \varphi^2}{\sqrt{1 + \frac{1}{x_n - 1}} + \varphi}\\ &= \frac{\frac{1}{x_n - 1} - \frac{1}{\varphi - 1}}{\sqrt{1 + \frac{1}{x_n - 1}} + \varphi}\\ &= \frac{\varphi - x_n}{\left (\varphi - 1\right )\left (\sqrt{x_n^2 - x_n} + \varphi\left(x_n - 1\right ) \right )}\\ \frac{x_{n+1} - \varphi}{\varphi - x_n} &= \frac{\varphi}{\left (x_{n+1} + \varphi\right )\left (\sqrt{x_n^2 - x_n} + \varphi\left(x_n - 1\right )\right )}\\ \end{align*} This result (which is true when the limits are substituted in) could be promising but I've come up on to a dead-end it seems.
Any help would be great.
After the edit, it should be simple. You have: $$x_{n+1} = f(x_n): \quad f(x) = \sqrt{1 + \dfrac{1}{\sqrt{1-\frac 1x}}},\,\, x_1 =\sqrt{2}.$$
Some, but not too terrible, calculus will show that $1 <x<f(x)$ if $x\in (1,\varphi).$ This tells us that $x_n$ is increasing. In fact, after removing square roots $f(x)>x$ is equivalent to: $$(x-1)^3(x+1)^2-x = (x^2-x-1)(x^3-x+1) < 0$$ which is easy to verify on $(1,\varphi).$
It's also trivially bounded too since because $f$ is a decreasing function: $$f(x_n)\leq f(x_1) = \sqrt{1 + \dfrac{1}{\sqrt{1-\frac{1}{\sqrt{2}}}}}.$$
So you have your convergence and the only thing left to do is to solve the equation: $$x = f(x)\iff (x^2-x-1)(x^3-x+1) = 0.$$ The unique positive solution is $\varphi$ and the problem follows.
EDIT:
The part of the above work that says $x_n$ is increasing is incorrect, as I mistakenly computed $f(x_1) < \varphi.$ But in reality, $f(x_1)\approx 1.68...$ and indeed the sequence oscillates. However, this makes it possible for a simpler proof using OP's work. One simply needs to have a lower bound on $x_n$ and in fact $x_n > 1$ will work since:
$$\frac{x_{n+1} - \varphi}{\varphi - x_n} = \frac{\varphi}{\left (x_{n+1} + \varphi\right )\left (\sqrt{x_n^2 - x_n} + \varphi\left(x_n - 1\right )\right )} < \dfrac{\varphi}{1+\varphi} = \varphi^{-1}<1.$$ This proves $x_n\to\varphi$ and in return OP's equation above proves the convergence rate is $\dfrac 14.$