Solve: $x = (x-\frac{1}{x}) ^ {1/9} + (1-\frac{1}{x})^{1/9}$ 
Solve: $$x = \left(x-\frac{1}{x}\right) ^ {1/9} + \left(1-\frac{1}{x}\right)^{1/9}$$  

Simplifying, $$x^{10/9} = (x^2-1)^{1/9}+(x-1)^{1/9}$$
I don't know how to start. Any hint will be helpful. 
 A: Seeing as OP was not satisfied with the numerical solution generated by Wolframalpha, there are ways to generate it with calculator alone.
As I see it, the problem is well posed for fixed-point iteration method. Let's start with $x \approx 2$ (it is easy to guess, as I will show later).
So we take $x_0=2$ for the first root. Then:
$$
x_1=\left(x_0−\frac{1}{x_0}\right)^{1/9}+\left(1−\frac{1}{x_0}\right)^{1/9}=
$$
$$
=\left(\frac{3}{2}\right)^{1/9}+\left(\frac{1}{2}\right)^{1/9}=1.97196...
$$
So we take $x_1=1.97196...$ and find $x_2$:
$$
x_2=\left(x_1−\frac{1}{x_1}\right)^{1/9}+\left(1−\frac{1}{x_1}\right)^{1/9}=1.9677...
$$
We are already very close to the solution (which is $1.966965...$) so the number of iterations we use depends on the precision we need.
The second root is tricky because it's too close to $1$. We can't take $x_0=1$ because then we'll get $x_1=0$, so we need to carefully choose the starting point. I will not elaborate further.

On the other hand, there is another way to approximate the second root. Since $x$ is really close to $1$ in this case, we can take $x=1+a$ and use the Taylor expansion:
$$
x = \left(1+a−\frac{1}{1+a}\right)^{1/9}+\left(1−\frac{1}{1+a}\right)^{1/9}
$$
$$
1+a \approx \left(2a\right)^{1/9}+\left(a\right)^{1/9}
$$
$$
a \approx \frac{1}{(2^{1/9}+1)^9}=0.00137...
$$
This answer is again very close to the accurate solution $1.00139...$. We can make it better by keeping 2 order terms in Taylor expansions.

As for $x \approx 2$ it is easy to show without calculator. We know that $a^{1/n} \approx 1$ for large $n$, unless $a$ is too small or too large. Let's eliminate these possibilities:
If $x \gg 1$ then from taking limits in the original equation we have:
$$
x \approx x^{1/9} \Rightarrow x \approx 1
$$
We arrived at a contradiction so $x$ can't be very large.
If $x \ll 1$ then from taking limits in the original equation we have:
$$
x \approx -2 \left(\frac{1}{x}\right)^{1/9} 
$$
Which doesn't even have real roots, so $x$ can't be very small.
From these we can show that both brackets have the same order, so if we take them both to be of the same order as $1$, we have:
$$
x \approx 1+1=2
$$
Note, that 'of the same order as $1$' doesn't mean they have to be especially close, for example $8^{1/9}=1.2599...$, $0.3^{1/9}=0.8748...$. Both values are close enough to $1$ after taking the root.
