Let f(x) be a bijective differentiable function and h(x) be its...

Since $h(x)$ is the inverse of $f(x)$, we have $f(h(x)) = x$.
From the given values:
$f(1) = 2 \implies h(2) = 1$
$f(3) = 9 \implies h(9) = 3$

Using the derivative of an inverse function, $h^{\prime}(x) = \dfrac{1}{f^{\prime}(h(x))}$.
At $x = 2$: $h^{\prime}(2) = \dfrac{1}{f^{\prime}(h(2))} = \dfrac{1}{f^{\prime}(1)} = \dfrac{1}{3}$
At $x = 9$: $h^{\prime}(9) = \dfrac{1}{f^{\prime}(h(9))} = \dfrac{1}{f^{\prime}(3)} = \dfrac{1}{1/2} = 2$

Now, differentiate $g(x) = (h(x))^2 \cdot h(x^3 + 1)$ using the product rule and chain rule:
$g^{\prime}(x) = \left[ 2h(x)h^{\prime}(x) \right] h(x^3 + 1) + (h(x))^2 \left[ h^{\prime}(x^3 + 1) \cdot 3x^2 \right]$

Substitute $x = 2$ into the derivative:
$g^{\prime}(2) = 2 h(2) h^{\prime}(2) h(9) + (h(2))^2 h^{\prime}(9) \cdot 3(2)^2$

Substitute the known values:
$g^{\prime}(2) = 2 (1) \left(\dfrac{1}{3}\right) (3) + (1)^2 (2) \cdot 12$
$g^{\prime}(2) = 2 + 24 = 26$

Answer: 26