We are tasked with finding the equation of the tangent line to the curve defined by:

$x^{2/3} + y^{2/3} = 5$

at the point $(-8, -1)$.

Steps to solve this problem:

1. Implicit Differentiation: We first differentiate both sides of the equation $x^{2/3} + y^{2/3} = 5$ with respect to $x$, keeping in mind that $y$ is a function of $x$ (implicit differentiation).

   Differentiating $x^{2/3}$ with respect to $x$: $\frac{d}{dx} \left( x^{2/3} \right) = \frac{2}{3}x^{-1/3}$

   Differentiating $y^{2/3}$ with respect to $x$: $\frac{d}{dx} \left( y^{2/3} \right) = \frac{2}{3}y^{-1/3} \cdot \frac{dy}{dx}$

2. Differentiation of the equation: Applying the product rule to the entire equation: $\frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3} \cdot \frac{dy}{dx} = 0$

3. Solve for $\frac{dy}{dx}$: Rearranging to isolate $\frac{dy}{dx}$: $\frac{2}{3}y^{-1/3} \cdot \frac{dy}{dx} = -\frac{2}{3}x^{-1/3}$ $y^{-1/3} \cdot \frac{dy}{dx} = -x^{-1/3}$ $\frac{dy}{dx} = -\frac{x^{-1/3}}{y^{-1/3}} = -\left(\frac{y}{x}\right)^{1/3}$

4. Substitute $(-8, -1)$ into the derivative: At the point $(-8, -1)$: $\frac{dy}{dx} = -\left(\frac{-1}{-8}\right)^{1/3} = -\left(\frac{1}{8}\right)^{1/3} = -\frac{1}{2}$

   So, the slope of the tangent line at $(-8, -1)$ is $-\frac{1}{2}$.

5. Equation of the tangent line: The equation of a line is given by: $y - y_1 = m(x - x_1)$ Substituting $m = -\frac{1}{2}$, $x_1 = -8$, and $y_1 = -1$: $y - (-1) = -\frac{1}{2}(x - (-8))$ $y + 1 = -\frac{1}{2}(x + 8)$ $y + 1 = -\frac{1}{2}x - 4$ $y = -\frac{1}{2}x - 5$

Thus, the equation of the tangent line is: $y = -\frac{1}{2}x - 5$

Let me know if you'd like further details or have any questions.

Here are 5 related questions to expand the concept:

1. What is implicit differentiation and when is it used?
2. How do you find the equation of a tangent line for parametric curves?
3. How does the behavior of the astroid change as you vary the value in its equation?
4. What is the geometric meaning of the slope at a given point on a curve?
5. How does the tangent line relate to the normal line?

Tip: When differentiating implicitly, remember that every time you differentiate a term involving $y$, you must multiply by $\frac{dy}{dx}$.