We'll first need to determine what the composite function f(g(x)) is equal to.
Start with the outer function f(x). Then replace every copy of x with g(x). Afterward, plug in g(x) = 3x-2
So we get the following:
[tex]f(x) = \sqrt{x^2-4}\\\\f(g(x)) = \sqrt{(g(x))^2-4}\\\\f(g(x)) = \sqrt{(3x-2)^2-4}\\\\f(g(x)) = \sqrt{9x^2-12x+4-4}\\\\f(g(x)) = \sqrt{9x^2-12x}\\\\[/tex]
Next, we apply the derivative
[tex]f(g(x)) = \sqrt{9x^2-12x}\\\\f(g(x)) = (9x^2-12x)^{1/2}\\\\f'(g(x)) = \frac{1}{2}(9x^2-12x)^{-1/2}*\frac{d}{dx}(9x^2-12x)\\\\f'(g(x)) = \frac{1}{2}(9x^2-12x)^{-1/2}*(18x-12)\\\\f'(g(x)) = \frac{9x-6}{\sqrt{9x^2-12x}}\\\\[/tex]
Don't forget about the chain rule. The chain rule is applied in step 3 of that block of steps shown above.
The last thing we do is plug in x = 3 and simplify.
[tex]f'(g(x)) = \frac{9x-6}{\sqrt{9x^2-12x}}\\\\f'(g(3)) = \frac{9*3-6}{\sqrt{9*3^2-12*3}}\\\\f'(g(3)) = \frac{21}{\sqrt{45}}\\\\f'(g(3)) = \frac{21}{\sqrt{9*5}}\\\\f'(g(3)) = \frac{21}{\sqrt{9}*\sqrt{5}}\\\\f'(g(3)) = \frac{21}{3*\sqrt{5}}\\\\f'(g(3)) = \frac{7}{\sqrt{5}}\\\\[/tex]
Side note: if you choose to rationalize the denominator, then you'll multiply both top and bottom by sqrt(5) to end up with [tex]f'(g(3)) = \frac{7\sqrt{5}}{5}\\\\[/tex]; however, your teacher has chosen not to rationalize the denominator.
