Let's work through the problem step by step, using logarithmic differentiation for the given function:

The function is: $y = \sqrt[3]{\frac{x^2(x+1)}{x^2-1}}$

Step 1: Take the natural logarithm of both sides

$\ln y = \ln \left( \sqrt[3]{\frac{x^2(x+1)}{x^2 - 1}} \right)$

Since $\ln(a^b) = b \ln a$, we can pull out the exponent $\frac{1}{3}$:

$\ln y = \frac{1}{3} \ln \left( \frac{x^2(x+1)}{x^2 - 1} \right)$

Step 2: Use the logarithmic property $\ln \left( \frac{a}{b} \right) = \ln a - \ln b$

$\ln y = \frac{1}{3} \left( \ln \left( x^2(x+1) \right) - \ln (x^2 - 1) \right)$

Step 3: Simplify using $\ln(ab) = \ln a + \ln b$

$\ln y = \frac{1}{3} \left( \ln x^2 + \ln(x+1) - \ln(x^2 - 1) \right)$

Now, simplify further: $\ln y = \frac{1}{3} \left( 2\ln x + \ln(x+1) - \ln(x^2 - 1) \right)$

Step 4: Differentiate both sides with respect to $x$

Using the chain rule:

$\frac{1}{y} \frac{dy}{dx} = \frac{1}{3} \left( \frac{2}{x} + \frac{1}{x+1} - \frac{2x}{x^2-1} \right)$

Step 5: Multiply both sides by $y$ to isolate $\frac{dy}{dx}$

We already know that $y = \sqrt[3]{\frac{x^2(x+1)}{x^2-1}}$, so we can substitute this back:

$\frac{dy}{dx} = \frac{1}{3} \left( \frac{2}{x} + \frac{1}{x+1} - \frac{2x}{x^2-1} \right) \cdot \sqrt[3]{\frac{x^2(x+1)}{x^2-1}}$

Step 6: Check the options

From this, the correct option corresponds to Option (b): $\frac{dy}{dx} = \frac{1}{3} \left( \frac{2}{x} + \frac{1}{x+1} - \frac{2x}{x^2-1} \right) \cdot \sqrt[3]{\frac{x^2(x+1)}{x^2-1}}$

Would you like further details on any step or explanations? Here are some related questions:

1. How do you differentiate logarithmic functions involving products and quotients?
2. What are the key rules of logarithmic differentiation?
3. How does the chain rule apply when differentiating logarithmic expressions?
4. Can you simplify the logarithmic differentiation process for other root functions?
5. How do you apply the chain rule for composite functions?

Tip: When dealing with functions that involve products, quotients, or roots, logarithmic differentiation simplifies the process by converting multiplicative relationships into additive ones, making differentiation easier.