We need to find the equation of the tangent line to the curve $y = \frac{-3e^x}{x}$ at the point $(1, -3e)$.

Steps:

1. Find the derivative: To determine the slope of the tangent line, we need to differentiate $y = \frac{-3e^x}{x}$ with respect to $x$.

   • Use the quotient rule: $\frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}$, where:
     • $f(x) = -3e^x$ and $g(x) = x$,
     • $f'(x) = -3e^x$ and $g'(x) = 1$.

2. Evaluate the derivative at $x = 1$: This will give the slope $m$ of the tangent line at the point $(1, -3e)$.

3. Find the equation of the tangent line: Once we have the slope $m$, we use the point-slope form of a line: $y - y_1 = m(x - x_1)$, where $(x_1, y_1) = (1, -3e)$.

Let me compute the derivative and the equation of the tangent line next.The derivative of the function $y = \frac{-3e^x}{x}$ at $x = 1$ gives a slope of 0. This means that the tangent line at the point $(1, -3e)$ is horizontal.

Thus, the equation of the tangent line is simply:

$y = -3e$

This is a constant line passing through $(1, -3e)$.

Would you like further details or have any questions? Here are some related problems you might want to explore:

1. How do we apply the quotient rule in other scenarios?
2. What happens if the function involves higher-order derivatives?
3. Can we find the tangent line for non-differentiable points?
4. How does the equation of the normal line relate to the tangent line?
5. Can we approximate the function using Taylor expansion around $x = 1$?

Tip: When the slope of a tangent line is zero, the line is horizontal, indicating that the function has a local extremum at that point (either a maximum or minimum).