but now you're on board the spaceship #369217211, and the captain is asking to solve tricky "ZP" (Z Planet) problem (you know what it means when captain is "asking"... that's an order):

(!!! Do Not use decimals/calculator. Present fractions in an exact format, for example, 7310773107 , 2295222952)

Now, find the slope of the tangent line to the curve

$$ x^2+3xy-3y^3=19 $$

at the point (1,−2)(1,-2).

  

!!! The challenge is that the easiest differentiation rule on Earth, ddxxn=nxn−1ddxxn=nxn-1 , on Z planet is "twisted" as the following: $$ \frac{d}{\differentialD x}x^{n}=n^2x^{n-1} $$  (ddxconst=0ddxconst=0 , and the other rules have not been changed).

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:
\[
\frac{dy}{dx} = -\frac{2}{33}.
\]

---

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.

but now you're on board the spaceship #369217211, and the captain is asking to solve tricky 'ZP' (Z Planet) problem: Find the slope of the tangent line to the curve x^2 + 3xy - 3y^3 = 19 at the point (1,−2), with the differentiation rule on Z planet: d/dx(x^n) = n^2x^(n-1).

This problem involves finding the slope of the tangent line to the curve x^2 + 3xy - 3y^3 = 19 at the point (1,-2), using a modified differentiation rule specific to Z Planet: d/dx(x^n) = n^2x^(n-1). Learn how to apply implicit differentiation and the product rule in this challenging calculus problem. Suitable for university-level calculus students.

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