How to show that $\sum_{n=1}^{\infty}{2^n\over (2^n-1)(2^{n+1}-1)(2^{n+2}-1)}={1\over 9}?$

Well done so far. Now shift the indices on the last two sums: $${1\over 3}\sum_{n=1}^{\infty}{1\over 2^n-1}-\sum_{n=1}^{\infty}{1\over 2^{n+1}-1}+{2\over 3}\sum_{n=1}^{\infty}{1\over 2^{n+2}-1}=\\ {1\over 3}\sum_{n=1}^{\infty}{1\over 2^n-1}-\sum_{n=2}^{\infty}{1\over 2^{n}-1}+{2\over 3}\sum_{n=3}^{\infty}{1\over 2^{n}-1}$$ and note that all the terms with $n \ge 3$ add to zero, so we just keep the first few terms: $$=\frac 13 \cdot \frac 1{2-1}+\frac 13\cdot \frac 1{4-1}-\frac 1{4-1}=\frac 19$$

You are almost done. Note that the three series on the LHS of your last line are convergent and they can be written as $${1\over 3}\sum_{n=1}^{\infty}{1\over 2^n-1}-\sum_{n=2}^{\infty}{1\over 2^{n}-1}+{2\over 3}\sum_{n=3}^{\infty}{1\over 2^{n}-1}$$ which is equal to $$ {1\over 3}\sum_{n=1}^{2}{1\over 2^n-1}-\sum_{n=2}^{2}{1\over 2^{n}-1}+\left(\frac{1}{3}-1+{2\over 3}\right)\sum_{n=3}^{\infty}{1\over 2^{n}-1}=\frac{1}{9}.$$

Hint: We can express

$$\frac{1}{2^n-1}=\sum_{k=1}^{\infty} \frac{1}{2^{kn}}.$$