Solving for $x$ is that much tough!!

$\begin{aligned} \sqrt x \left(9^{\sqrt{x^2-3}} - 3^{\sqrt{x^2-3}} \right) & = 3^{2\sqrt{x^2-3}+1} - 3^{\sqrt{x^2-3}+1} + 6\sqrt x - 18 \\ \sqrt x \left(3^{2\sqrt{x^2-3}} - 3^{\sqrt{x^2-3}} \right) & = 3 \cdot 3^{2\sqrt{x^2-3}} - 3 \cdot 3^{\sqrt{x^2-3}} + 6\sqrt x - 18 \\ \left(\sqrt x - 3\right) \left(3^{2\sqrt{x^2-3}} - 3^{\sqrt{x^2-3}} \right) - 6 \left(\sqrt x - 3\right) & = 0 \\ \left(\sqrt x - 3\right) \left(3^{2\sqrt{x^2-3}} - 3^{\sqrt{x^2-3}} - 6 \right) & = 0 \end{aligned}$

$\implies \begin{cases} \sqrt x - 3 = 0 & \implies x = 9 \\ 3^{2\sqrt{x^2-3}} - 3^{\sqrt{x^2-3}} - 6 = 0 & \implies \left(3^{\sqrt{x^2-3}} +2 \right) \left(3^{\sqrt{x^2-3}}-3 \right) = 0 & \implies 3^{\sqrt{x^2-3}} = 3 & \implies x = 2 \end{cases}$

The largest value of $x$ is $\boxed{9}$ .

Not a solution but a comment: The posted answer is $x = 9$ , but $x = 2$ also works.

Simple enough; let 3^\sqrt{x^2-3}=a

The equation becomes

$\sqrt{x}(a^2-a)=3a^2-3a+6\sqrt{x}-18$

Moving $6\sqrt{x}$ over, factor the left side by grouping and factoring out $3$ on the right side:

$\sqrt{x}(a^2-a-6)=3(a^2-a-6)$

From this we can see that $\sqrt{x}=3$ , thus $x=\boxed{9}$

So, is the real answer finally x=2 or x=9?

It was just a matter of correct factorization, good problem.