We define $f_0(x) = 8$ . Then the recursive relation holds for $n=0$ , as well.
Since $f_n (x) \ge 0$ for all nonnegative integers $n$ , it suffices to consider nonnegative values of $x$ .
We claim that the following set of relations hold true for all natural numbers $n$ and nonnegative reals $x$ : $$\begin{align*} f_n(x) &< 2x \text{ if }x>4 ; \\ f_n(x) &= 2x \text{ if }x=4 ; \\ f_n(x) &> 2x \text{ if }x<4 . \end{align*}$$ To prove this claim, we induct on $n$ . The statement evidently holds for our base case, $n=0$ .
Now, suppose the claim holds for $n$ . Then $$\begin{align*} f_{n+1}(x) &= \sqrt{x^2 + 6f_n(x)} < \sqrt{x^2+12x} < \sqrt{4x^2} = 2x, \text{ if } x>4 ; \\ f_{n+1}(x) &= \sqrt{x^2 + 6f_n(x)} = \sqrt{x^2 + 12x} = \sqrt{4x^2} = 2x, \text{ if } x=4 ; \\ f_{n+1}(x) &= \sqrt{x^2 + 6f_n(x)} > \sqrt{x^2+12x} > \sqrt{4x^2} = 2x, \text{ if } x<4 . \end{align*}$$ The claim therefore holds by induction. It then follows that for all nonnegative integers $n$ , $x=4$ is the unique solution to the equation $f_n(x) = 2x$ . $\blacksquare$
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