By the chain rule,
[tex]g(x) = f(4 - x^2) \\\\ \implies g'(x) = -2x f'(4 - x^2) \\\\ \implies g''(x) = -2 f'(4 - x^2) + 4x^2 f''(4 - x^2)[/tex]
Observe that
[tex]4 - x^2 = 1 \implies x^2 = 3 \implies x = \pm\sqrt3[/tex]
Then
[tex]g''(\sqrt3) = -2 f'(1) + 4\left(\sqrt3\right)^2 f''(1) = -2(-5) + 4(3)(-1) = \boxed{-2}[/tex]