Suppose that $$ 3 x^{2} y^{2}+2 x-4 y=1 $$. Find the derivative $$ \frac{d y}{d x} $$ at the point $$ (1,1) $$

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To find the derivative $$\frac{dy}{dx}$$ for the equation $$3x^{2}y^{2} + 2x - 4y = 1$$ at the point $$(1, 1)$$, we can use implicit differentiation. Here's a step-by-step solution:

1. **Differentiate both sides of the equation with respect to $$x$$:**

$$
   \frac{d}{dx}(3x^{2}y^{2}) + \frac{d}{dx}(2x) - \frac{d}{dx}(4y) = \frac{d}{dx}(1)
   $$

2. **Apply the product rule and chain rule where necessary:**

- For $$3x^{2}y^{2}$$:
     $$
     \frac{d}{dx}(3x^{2}y^{2}) = 3 \left(2x y^{2} + x^{2} \cdot 2y \frac{dy}{dx}\right) = 6x y^{2} + 6x^{2} y \frac{dy}{dx}
     $$
   
   - For $$2x$$:
     $$
     \frac{d}{dx}(2x) = 2
     $$
   
   - For $$-4y$$:
     $$
     \frac{d}{dx}(-4y) = -4 \frac{dy}{dx}
     $$
   
   - The derivative of a constant is zero:
     $$
     \frac{d}{dx}(1) = 0
     $$

3. **Combine all the differentiated components:**

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

4. **Solve for $$\frac{dy}{dx}$$:**

$$
   (6x^{2} y - 4) \frac{dy}{dx} = -6x y^{2} - 2
   $$
   
   $$
   \frac{dy}{dx} = \frac{-6x y^{2} - 2}{6x^{2} y - 4}
   $$

5. **Evaluate at the point $$(1, 1)$$:**

$$
   \frac{dy}{dx}\Bigg|_{(1,1)} = \frac{-6(1)(1)^{2} - 2}{6(1)^{2}(1) - 4} = \frac{-6 - 2}{6 - 4} = \frac{-8}{2} = -4
   $$

**Final Answer:**  
At the point $$(1, 1)$$, the derivative $$\frac{dy}{dx}$$ is $$-4$$.
At the point (1,1), the derivative $$\frac{dy}{dx}$$ is -4.

Suppose that . Find the derivative at the point

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