<s>[INST] On the equation $\sqrt{x^2-9}=\frac{2(x+3)}{(x-3)^2}-x$ I'm trying to solve the equation
$$\sqrt{x^2-9}=\frac{2(x+3)}{(x-3)^2}-x \tag{1}$$
Attempt: I have rewritten the equation as
\begin{align*} 
\sqrt{x^2-9} = \frac{2(x+3)}{\left (x-3 \right )^2}-x &\Leftrightarrow \sqrt{x^2-9} = \frac{2(x+3)-x\left ( x-3 \right )^2}{\left ( x-3 \right )^2} \\ 
&\Leftrightarrow \sqrt{x^2-9} = - \frac{x^3-6x^2+7x-6}{\left ( x-3 \right )^2} \\ &\Leftrightarrow \sqrt{x^2-9} = \frac{2}{x-3} -x + \frac{12}{\left ( x-3 \right )^2} 
\end{align*}
This seems manageable (?) but I do not know how to proceed. The solution is $x=8-\sqrt{13}$ as suggested by Mr. Wolfy. I have no idea how to get it.
In the mean time if we square $(1)$ all 6th powers are simplified and we are left with 
$$5x^4-96x^3+526x^2-105x+765=0$$
Trying to factorising it we get with some luck that
$$(x^2-16x+51)(5x^2-16x+15) =0$$
and we have to solve this. This is easy but we have to take into account the restriction 
$$ -x^3+6x^2-7x+6\geq 0 $$
 [/INST] Difference of two squares identity and its Conjugates can help:
$\sqrt{x^2-9} = \frac{2(x+3)}{(x-3)^2} - x$
$(x-3)^2(\sqrt{x^2-9} + x) = 2(x+3)$
$(x-3)^2(\sqrt{x^2-9} + x)(\sqrt{x^2-9} - x) = 2(x+3)(\sqrt{x^2-9} - x)$
$(x-3)^2(x^2-9 - x^2) = 2(x+3)(\sqrt{x^2-9} - x)$
$-9(x-3)^2 = 2(x+3)(\sqrt{x^2-9} - x)$
$\frac{-9(x-3)^2}{2(x+3)} = \sqrt{x^2-9} - x$
$\frac{-9(x-3)^2}{2(x+3)} + x = \sqrt{x^2-9}$
We achieve new equation for $\sqrt{x^2-9}, \ $ and so we can equal this one to the first equation:
$\frac{-9(x-3)^2}{2(x+3)} + x = \sqrt{x^2-9} = \frac{2(x+3)}{(x-3)^2} - x$
$\frac{-9(x-3)^2}{2(x+3)} + x = \frac{2(x+3)}{(x-3)^2} - x$
Now we get ride of radicals, so we can solve it without any restriction caused by squaring.
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