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The problem is asking to find the slope of the tangent line to the implicitly defined curve at a specific point. The equation given is:

\[
-3x^2 + 4xy + 4y^3 = -99
\]

at the point \((-3, -3)\).

### Steps to solve:
1. **Implicit Differentiation**: 
   To find the slope of the tangent line, we need to implicitly differentiate the given equation with respect to \(x\), treating \(y\) as a function of \(x\), i.e., \(y = y(x)\).

\[
\frac{d}{dx} \left(-3x^2 + 4xy + 4y^3 \right) = \frac{d}{dx}(-99)
\]

Differentiating each term:

- For \(-3x^2\), we get:
\[
\frac{d}{dx}(-3x^2) = -6x
\]

- For \(4xy\), we use the product rule:
\[
\frac{d}{dx}(4xy) = 4 \left( \frac{d}{dx}(x) \cdot y + x \cdot \frac{dy}{dx} \right) = 4(y + x \frac{dy}{dx})
\]

- For \(4y^3\), we use the chain rule:
\[
\frac{d}{dx}(4y^3) = 4 \cdot 3y^2 \cdot \frac{dy}{dx} = 12y^2 \frac{dy}{dx}
\]

Now, the derivative of the constant \(-99\) is \(0\), so putting it all together:

\[
-6x + 4(y + x \frac{dy}{dx}) + 12y^2 \frac{dy}{dx} = 0
\]

2. **Solving for \(\frac{dy}{dx}\)**:
Now, solve for \(\frac{dy}{dx}\) (which represents the slope of the tangent line):

\[
-6x + 4y + 4x \frac{dy}{dx} + 12y^2 \frac{dy}{dx} = 0
\]

Isolate \(\frac{dy}{dx}\):

\[
4x \frac{dy}{dx} + 12y^2 \frac{dy}{dx} = 6x - 4y
\]

Factor out \(\frac{dy}{dx}\):

\[
\frac{dy}{dx} (4x + 12y^2) = 6x - 4y
\]

Solve for \(\frac{dy}{dx}\):

\[
\frac{dy}{dx} = \frac{6x - 4y}{4x + 12y^2}
\]

3. **Substitute the point \((-3, -3)\)**:
Now, substitute \(x = -3\) and \(y = -3\) into the equation:

\[
\frac{dy}{dx} = \frac{6(-3) - 4(-3)}{4(-3) + 12(-3)^2}
\]

Simplify the numerator:

\[
6(-3) - 4(-3) = -18 + 12 = -6
\]

Simplify the denominator:

\[
4(-3) + 12(-3)^2 = -12 + 12(9) = -12 + 108 = 96
\]

Thus, the slope is:

\[
\frac{dy}{dx} = \frac{-6}{96} = -\frac{1}{16}
\]

### Final Answer:
The slope of the tangent line at the point \((-3, -3)\) is:

\[
m = -\frac{1}{16}
\]

Would you like further clarification on any part of the solution? 

Here are 5 related questions for further exploration:
1. How do you apply implicit differentiation to other types of functions?
2. What is the product rule, and how is it used in implicit differentiation?
3. How do you determine if a function is implicitly defined or explicitly defined?
4. How can you find the equation of the tangent line, given the slope?
5. What happens when you differentiate higher-order terms like \(y^3\)?

**Tip**: When using implicit differentiation, always remember to apply the chain rule to any \(y\) terms, as they are functions of \(x\).

Find the slope of the tangent line drawn to the graph of -3x^2 + 4xy + 4y^3 = -99 at the point (-3, -3). Enter exact values.

Learn how to find the slope of the tangent line to the implicitly defined curve -3x^2 + 4xy + 4y^3 = -99 at the point (-3, -3) using implicit differentiation. Suitable for students in Grades 11-12.

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