If $f(x)=\sqrt{x^2-4}$ and $g(x)=3x-2$, then the derivative of $f(g(x))$ at $x=3$ is

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Use the chain rule to find the expression for the derivative:

$$\frac{d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$$

At $x=3$,

$$f'(g(3))\cdot g'(3)$$

Obtaining the equations for $f'$ and $g'$:

Using our equations:

$$g(3)=7$$ $$g'(3)=3$$

Putting it all together:

$$f'(g(3))\cdot g'(3)$$ $$= f'(7)\cdot 3$$ $$= \frac{7}{\sqrt{45}} \cdot 3$$ $$= \frac{7}{3\sqrt{5}}\cdot 3$$ $$= \boxed{\frac{7}{\sqrt{5}}}$$