The problem asks for the slope of the tangent line to the curve $x^2 + 3xy - 3y^3 = 19$ at the point $(1, -2)$, with a modified differentiation rule on the Z planet.

Step 1: Implicit Differentiation

We need to find $\frac{dy}{dx}$ using implicit differentiation on the equation $x^2 + 3xy - 3y^3 = 19$, applying the modified differentiation rule: $\frac{d}{dx}x^n = n^2x^{n-1}.$

Differentiate each term:

1. First term: $x^2$
   Using the modified rule: $\frac{d}{dx}(x^2) = 2^2x^{2-1} = 4x.$

2. Second term: $3xy$
   We need to use the product rule, which remains unchanged:
   $\frac{d}{dx}(3xy) = 3\left( \frac{d}{dx}(x)y + x\frac{dy}{dx} \right).$ First, apply the Z planet rule to $\frac{d}{dx}(x)$: $\frac{d}{dx}(x) = 1^2x^{1-1} = 1.$ Thus: $\frac{d}{dx}(3xy) = 3(1 \cdot y + x \cdot \frac{dy}{dx}) = 3(y + x\frac{dy}{dx}).$

3. Third term: $-3y^3$
   Since $y$ is a function of $x$, we apply the chain rule: $\frac{d}{dx}(-3y^3) = -3 \cdot 3y^2 \frac{dy}{dx} = -9y^2 \frac{dy}{dx}.$

Step 2: Combine the Results

Now we combine all the differentiated terms: $4x + 3(y + x\frac{dy}{dx}) - 9y^2 \frac{dy}{dx} = 0.$ Simplify: $4x + 3y + 3x\frac{dy}{dx} - 9y^2 \frac{dy}{dx} = 0.$

Step 3: Solve for $\frac{dy}{dx}$

Rearrange the terms to isolate $\frac{dy}{dx}$: $3x\frac{dy}{dx} - 9y^2 \frac{dy}{dx} = -4x - 3y,$ $\frac{dy}{dx}(3x - 9y^2) = -4x - 3y,$ $\frac{dy}{dx} = \frac{-4x - 3y}{3x - 9y^2}.$

Step 4: Substitute the Point $(1, -2)$

Now substitute $x = 1$ and $y = -2$ into the expression for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{-4(1) - 3(-2)}{3(1) - 9(-2)^2},$ $\frac{dy}{dx} = \frac{-4 + 6}{3 - 9(4)},$ $\frac{dy}{dx} = \frac{2}{3 - 36} = \frac{2}{-33}.$

Thus, the slope of the tangent line at the point $(1, -2)$ is $\frac{2}{-33}$, or simply: $\frac{dy}{dx} = -\frac{2}{33}.$

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Would you like more details or have any questions?

Here are 5 related questions to extend your learning:

1. How does the Z planet's differentiation rule differ from Earth's?
2. What changes if the curve involves trigonometric functions under Z planet rules?
3. How would implicit differentiation work for curves of higher degrees on Z planet?
4. Can you derive the normal line equation at the same point on Earth, using Earth's rules?
5. How does the product rule remain unaffected under Z planet differentiation?

Tip: When solving implicit differentiation problems, keep track of both direct and implicit derivatives to avoid confusion.