Let k and m be positive real numbers such that the function $$f(x) = \begin{cases} 3x^2 + k\sqrt{x + 1}, & 0 < x < 1 \\ mx^2 + k^2, & x \geq 1 \end{cases}$$ is differentiable for all $$x > 0$$. Then $$\frac{8f'(8)}{f'(\frac{1}{8})}$$ is equal to _____.

We are given a piecewise function:

$$f(x) = \begin{cases} 3x^2 + k\sqrt{x + 1}, & 0 < x < 1 \\ mx^2 + k^2, & x \geq 1 \end{cases}$$

where $$k$$ and $$m$$ are positive real numbers, and $$f(x)$$ is differentiable for all $$x > 0$$.

For $$f$$ to be differentiable at $$x = 1$$, it must first be continuous there. Equating the left-hand and right-hand limits:

$$3(1)^2 + k\sqrt{1 + 1} = m(1)^2 + k^2$$

$$3 + k\sqrt{2} = m + k^2 \quad \cdots (i)$$

The derivatives of each piece are:

For $$0 < x < 1$$: $$f'(x) = 6x + \dfrac{k}{2\sqrt{x+1}}$$

For $$x \geq 1$$: $$f'(x) = 2mx$$

Equating derivatives at $$x = 1$$:

$$6 + \dfrac{k}{2\sqrt{2}} = 2m \quad \cdots (ii)$$

From equation (ii): $$m = 3 + \dfrac{k}{4\sqrt{2}}$$

Substituting into equation (i):

$$3 + k\sqrt{2} = 3 + \dfrac{k}{4\sqrt{2}} + k^2$$

$$k\sqrt{2} - \dfrac{k}{4\sqrt{2}} = k^2$$

$$k\left(\sqrt{2} - \dfrac{1}{4\sqrt{2}}\right) = k^2$$

Since $$k > 0$$, we divide both sides by $$k$$:

$$k = \sqrt{2} - \dfrac{1}{4\sqrt{2}} = \dfrac{8 - 1}{4\sqrt{2}} = \dfrac{7}{4\sqrt{2}} = \dfrac{7\sqrt{2}}{8}$$

Then from (ii): $$2m = 6 + \dfrac{7\sqrt{2}}{8 \cdot 2\sqrt{2}} = 6 + \dfrac{7}{16} = \dfrac{103}{16}$$

$$m = \dfrac{103}{32}$$

Since $$8 \geq 1$$, we use $$f'(x) = 2mx$$:

$$f'(8) = 2 \cdot \dfrac{103}{32} \cdot 8 = \dfrac{103}{2}$$

Since $$0 < \dfrac{1}{8} < 1$$, we use $$f'(x) = 6x + \dfrac{k}{2\sqrt{x+1}}$$:

$$f'\!\left(\dfrac{1}{8}\right) = \dfrac{3}{4} + \dfrac{\frac{7\sqrt{2}}{8}}{2\sqrt{\frac{9}{8}}} = \dfrac{3}{4} + \dfrac{\frac{7\sqrt{2}}{8}}{\frac{3}{\sqrt{2}}} = \dfrac{3}{4} + \dfrac{7\sqrt{2} \cdot \sqrt{2}}{8 \cdot 3} = \dfrac{3}{4} + \dfrac{7}{12} = \dfrac{4}{3}$$

$$\dfrac{8f'(8)}{f'\!\left(\dfrac{1}{8}\right)} = \dfrac{8 \cdot \dfrac{103}{2}}{\dfrac{4}{3}} = \dfrac{412}{\dfrac{4}{3}} = 412 \times \dfrac{3}{4} = \boxed{309}$$