If $f(x) = \frac{x}{x - \frac{1}{f(x)}}$, then what is the fixed point of $f(f(x))$?

Question

Let $x \in \mathbb{R}$ and suppose that $x>1$.

Here is my question:

What is the fixed point of $f(f(x))$ if $$f(x) = \dfrac{x}{x - \dfrac{1}{f(x)}}?$$

My Attempt

$$f(f(x)) = \dfrac{x}{x - \dfrac{1}{\dfrac{x}{x - \dfrac{1}{f(x)}}}} = \dfrac{x}{x - \dfrac{x - \dfrac{1}{f(x)}}{x}} = \dfrac{x^2}{x^2 - x + \dfrac{1}{f(x)}}$$ $$f(f(x))=f(x) \iff f(x) = \dfrac{x}{x - \dfrac{1}{f(x)}}$$

Solving for $f(x)$, we obtain $$f(x) = \frac{xf(x)}{xf(x) - 1}$$ so that $$x(f(x))^2 - f(x) = xf(x).$$ It follows that $$x(f(x))^2 = (x+1)f(x).$$ Since $f(x) \neq 0$, we can cancel $f(x)$ from both sides, and thereby get $$f(x)=\frac{x+1}{x}=1+\frac{1}{x}.$$

We therefore conclude that the fixed point of $f(f(x))$ is $$f(x) = \frac{x+1}{x}.$$

Is my interpretation of the word fixed point in the context of this question correct?

Update (September 25 2016 - 12:32 AM) Per a comment from b00n heT, I should not use the term "fixed point". A follow-up question of mine would then be: What terminology should I use?