<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] Hint:
Let $\sqrt{x^2-9}=3\tan2t\ge0,0\le2t<\dfrac\pi2$
$x=3\sec2t$
$$3(\tan2t+\sec2t)=\dfrac{6(\sec2t+1)}{9(\sec2t-1)^2}$$
$$3(\sin2t+1)=\dfrac{2(1+\cos2t)}{3(1-\cos2t)^2}$$
Set $\tan t=u$
$$\dfrac{9(1+u)^2}{1+u^2}=\dfrac{4(1+u^2)^2}{4(1+u^2)u^4}$$
$$\iff9(1+u)^2u^4=(1+u^2)^2$$
As $u=\tan t\ge0$
$$\implies3(1+u)u^2=1+u^2$$
$$3u^3+2u^2-1=0$$
Use Cardano's method
</s>