To find the second derivative of $f(x) = \sqrt[3]{(x^2 + 7)^2}$, we need to go through the following steps:

1. Rewrite the Function in Power Form:

   Rewrite $f(x)$ in a form that makes differentiation easier: $f(x) = (x^2 + 7)^{\frac{2}{3}}$

2. Find the First Derivative, $f'(x)$:

   Using the chain rule, we differentiate $f(x) = (x^2 + 7)^{\frac{2}{3}}$: $f'(x) = \frac{2}{3}(x^2 + 7)^{\frac{-1}{3}} \cdot (2x)$ Simplifying, we get: $f'(x) = \frac{4x}{3 (x^2 + 7)^{\frac{1}{3}}}$

3. Find the Second Derivative, $f''(x)$:

   Now, we differentiate $f'(x)$ again with respect to $x$: $f''(x) = \frac{d}{dx} \left( \frac{4x}{3 (x^2 + 7)^{\frac{1}{3}}} \right)$

   We apply the quotient rule here: $f''(x) = \frac{(3(x^2 + 7)^{\frac{1}{3}} \cdot 4) - (4x \cdot \frac{d}{dx}((x^2 + 7)^{\frac{1}{3}}))}{\left( 3(x^2 + 7)^{\frac{1}{3}} \right)^2}$

   Now we calculate $\frac{d}{dx}((x^2 + 7)^{\frac{1}{3}})$: $\frac{d}{dx}((x^2 + 7)^{\frac{1}{3}}) = \frac{1}{3}(x^2 + 7)^{-\frac{2}{3}} \cdot (2x) = \frac{2x}{3(x^2 + 7)^{\frac{2}{3}}}$

   Substitute this back in: $f''(x) = \frac{4 \cdot 3 (x^2 + 7)^{\frac{1}{3}} - 4x \cdot \frac{2x}{3(x^2 + 7)^{\frac{2}{3}}}}{9(x^2 + 7)^{\frac{2}{3}}}$

   Simplify the expression further: $f''(x) = \frac{12(x^2 + 7)^{\frac{1}{3}} - \frac{8x^2}{(x^2 + 7)^{\frac{2}{3}}}}{9(x^2 + 7)^{\frac{2}{3}}}$

   Finally, we can combine terms: $f''(x) = \frac{12(x^2 + 7) - 8x^2}{9(x^2 + 7)^{\frac{5}{3}}}$

So, $f''(x) = \frac{12x^2 + 84 - 8x^2}{9(x^2 + 7)^{\frac{5}{3}}} = \frac{4x^2 + 84}{9(x^2 + 7)^{\frac{5}{3}}}$

This is the second derivative, $f''(x)$.

Would you like more clarification on any step? Here are some related questions:

1. How is the chain rule used in differentiation?
2. What is the power rule in differentiation?
3. How does the quotient rule apply in finding second derivatives?
4. How do we handle derivatives of fractional exponents?
5. What are some practical applications of second derivatives?

Tip: Always simplify expressions as much as possible before differentiating again.