To locate any local maximum or minimum points of the function $f(x) = \sqrt[3]{x^2 + 8}$ using the Second Derivative Test, follow these steps:

1. First Derivative: To find the critical points, we need to compute the first derivative of the function and set it to zero.

   $f(x) = (x^2 + 8)^{\frac{1}{3}}$

   Using the chain rule:

   $f'(x) = \frac{d}{dx} \left( (x^2 + 8)^{\frac{1}{3}} \right)$

   Let $u = x^2 + 8$. Then $f(x) = u^{\frac{1}{3}}$, and by the chain rule:

   $f'(x) = \frac{d}{du} \left( u^{\frac{1}{3}} \right) \cdot \frac{du}{dx}$

   $f'(x) = \frac{1}{3} u^{-\frac{2}{3}} \cdot (2x) = \frac{1}{3} (x^2 + 8)^{-\frac{2}{3}} \cdot 2x$

   Simplifying:

   $f'(x) = \frac{2x}{3 (x^2 + 8)^{\frac{2}{3}}}$

2. Set the First Derivative to Zero: To find the critical points, set $f'(x) = 0$:

   $\frac{2x}{3 (x^2 + 8)^{\frac{2}{3}}} = 0$

   This equation is zero when the numerator is zero:

   $2x = 0 \implies x = 0$

   So, $x = 0$ is a critical point.

3. Second Derivative: Next, compute the second derivative to apply the Second Derivative Test.

   $f'(x) = \frac{2x}{3 (x^2 + 8)^{\frac{2}{3}}}$

   We need the derivative of $f'(x)$. Let’s denote:

   $g(x) = 2x \quad \text{and} \quad h(x) = 3 (x^2 + 8)^{\frac{2}{3}}$

   So, $f'(x) = \frac{g(x)}{h(x)}$, and use the quotient rule for the second derivative:

   $f''(x) = \frac{g'(x) h(x) - g(x) h'(x)}{(h(x))^2}$

   Calculate $g'(x)$ and $h'(x)$:

   $g'(x) = 2$

   $h(x) = 3 (x^2 + 8)^{\frac{2}{3}}$

   Using the chain rule:

   $h'(x) = 3 \cdot \frac{2}{3} (x^2 + 8)^{-\frac{1}{3}} \cdot 2x = 2 \cdot 2x (x^2 + 8)^{-\frac{1}{3}} = \frac{4x}{(x^2 + 8)^{\frac{1}{3}}}$

   Substitute back into the quotient rule:

   $f''(x) = \frac{2 \cdot 3 (x^2 + 8)^{\frac{2}{3}} - 2x \cdot \frac{4x}{(x^2 + 8)^{\frac{1}{3}}}}{(3 (x^2 + 8)^{\frac{2}{3}})^2}$

   Simplifying the numerator:

   $= \frac{6 (x^2 + 8)^{\frac{2}{3}} - \frac{8x^2}{(x^2 + 8)^{\frac{1}{3}}}}{9 (x^2 + 8)^{\frac{4}{3}}}$

   Combine terms in the numerator:

   $= \frac{6 (x^2 + 8) - 8x^2}{9 (x^2 + 8)^{\frac{4}{3}}} = \frac{48 - 2x^2}{9 (x^2 + 8)^{\frac{4}{3}}}$

   Evaluate the second derivative at the critical point $x = 0$:

   $f''(0) = \frac{48 - 2(0)^2}{9 (0^2 + 8)^{\frac{4}{3}}} = \frac{48}{9 \cdot 8^{\frac{4}{3}}} = \frac{48}{72} = \frac{2}{3}$

4. Second Derivative Test: Since $f''(0) > 0$, the function $f(x)$ has a local minimum at $x = 0$.

So, the function $f(x) = \sqrt[3]{x^2 + 8}$ has a local minimum at $x = 0$.

Would you like more details on any step or further explanation?

Here are some related questions:

1. What is the general form of the chain rule?
2. How do you apply the quotient rule for derivatives?
3. What are critical points and how are they identified?
4. How do you interpret the second derivative in the context of concavity?
5. What are other methods for finding local extrema?
6. Can you explain the significance of the Second Derivative Test?
7. How do you solve the equation $\frac{2x}{3 (x^2 + 8)^{2/3}} = 0$ in general?
8. What happens if the second derivative at a critical point is zero?

Tip: When using the chain rule, always identify the inner and outer functions clearly to avoid errors.