If $f\left(x-\frac{2}{x}\right) = \sqrt{x-1}$, then what is the value of $f'(1)$ Find $f'(1)$ if $$f\left(x-\frac{2}{x}\right) = \sqrt{x-1}$$ 


My attempt at the question:

Let $(x-\dfrac{2}{x})$ be $g(x)$
Then $$f(g(x)) = \sqrt{x-1} $$
Differentiating with respect to x: 
$$f'(g(x))\cdot g'(x) = \frac{1}{2\sqrt{x-1}} $$
Therefore

$$f'(g(x)) = \frac{1}{2(g'(x))\sqrt{x-1}}  $$

Finding the value of $x$ for which $g(x) = 1$ : $ x=( -1) , x=2$
But as $x\neq (-1)$, as $\sqrt{x-1}$ becomes indeterminant, substitute x = 2.
we get: $$f'(1) = \frac13 $$ 
Which is not the correct answer. The correct answer is supposedly $1$. Need some help as to why my method is wrong.
The answer is indeed $\frac13$. To check this, let's go another way: set $g=x-\frac2x$, as you did. Then express $x$ through $g$: 
$$x=\frac12\left(g\pm\sqrt{g^2+8}\right).$$

Then

$$f(g)=\sqrt{\frac12\left(g\pm\sqrt{g^2+8}\right)-1}.$$

Taking its derivative, we can find that it's real at $g=1$ for $\pm\to+$, and we get
$$f'(1)=\frac13.$$

(The other solution with $\pm\to-$ is $-\frac{i}{6\sqrt2}$).

Why does your source say the answer is $1$? Maybe it's just a mistake of evaluating $f(1)$ instead of $f'(1)$, since $f(1)=1$.
Let us assume ${\left(x-\frac{2}{x}\right) = 1, \text{the part inside}}$

It is found that $x = 2$ or $x = -1$

Taking the positive value of $x$

Therefore $f(1) = \sqrt {x-1}$ and 
Thus $$f'(1) = \frac {1}{2}\cdot(x-1)^{\frac{1}{2}-1},$$ $ \text{1 for the value inside the function}$
$$\Rightarrow \frac{1}{2}\cdot(x-1)^{\frac{-1}{2}}$$
Putting the positive value $2$, we get 
$f'(1) = \frac{1}{2}$