To solve this problem, we need to implicitly differentiate the given equation with respect to $x$ and then solve for $\frac{dy}{dx}$.

The given equation is:

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

Step 1: Implicit Differentiation

We'll differentiate both sides of the equation with respect to $x$, remembering that $y$ is a function of $x$, so we must use the chain rule when differentiating terms involving $y$.

Left-hand side:

• The derivative of $-y^3$ with respect to $x$ is: $-3y^2 \frac{dy}{dx}$ (by the chain rule).
• The derivative of $x^3$ with respect to $x$ is: $3x^2$
• The derivative of $x^2$ with respect to $x$ is: $2x$

So, the derivative of the left-hand side is: $-3y^2 \frac{dy}{dx} + 3x^2 + 2x$

Right-hand side:

• The derivative of $-3y$ with respect to $x$ is: $-3 \frac{dy}{dx}$ (using the chain rule).
• The derivative of $-2y^2$ with respect to $x$ is: $-4y \frac{dy}{dx}$ (using the chain rule).

So, the derivative of the right-hand side is: $-3 \frac{dy}{dx} - 4y \frac{dy}{dx}$

Step 2: Setting up the equation

Now, we equate the derivatives of both sides:

$-3y^2 \frac{dy}{dx} + 3x^2 + 2x = -3 \frac{dy}{dx} - 4y \frac{dy}{dx}$

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

We'll gather all terms involving $\frac{dy}{dx}$ on one side:

$-3y^2 \frac{dy}{dx} + 3 \frac{dy}{dx} + 4y \frac{dy}{dx} = 3x^2 + 2x$

Factor out $\frac{dy}{dx}$:

$\frac{dy}{dx} (-3y^2 + 3 + 4y) = 3x^2 + 2x$

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

$\frac{dy}{dx} = \frac{3x^2 + 2x}{3 + 4y - 3y^2}$

Step 4: Evaluate at the point $(-1, 3)$

Now substitute $x = -1$ and $y = 3$ into the expression for $\frac{dy}{dx}$:

• For the numerator: $3(-1)^2 + 2(-1) = 3(1) - 2 = 1$

• For the denominator: $3 + 4(3) - 3(3)^2 = 3 + 12 - 27 = -12$

Thus:

$\frac{dy}{dx} = \frac{1}{-12} = -\frac{1}{12}$

Final Answer:

$\frac{dy}{dx} = -\frac{1}{12} \text{ at the point } (-1, 3)$

Would you like more details or have any questions? Here are some related questions to explore:

1. What is implicit differentiation, and why do we use it?
2. How does the chain rule apply in implicit differentiation?
3. What are some examples where implicit differentiation is necessary?
4. Can you differentiate a different equation implicitly and solve for $\frac{dy}{dx}$?
5. How would the answer change if the equation had additional terms?

Tip: Always keep track of where you apply the chain rule when differentiating implicitly!