To solve for $\frac{dy}{dx}$ at $x = 1$, given the equation:

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

we'll need to differentiate both sides of the equation with respect to $x$ using implicit differentiation.

Step 1: Differentiate each term with respect to $x$

1. For $3x^2$, the derivative is $6x$.
2. For $2xy$, we use the product rule: $\frac{d}{dx}(2xy) = 2 \left(\frac{dx}{dx} \cdot y + x \cdot \frac{dy}{dx}\right) = 2y + 2x \frac{dy}{dx}$
3. For $y^2$, we use the chain rule: $\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}$

Step 2: Differentiate the entire equation

Differentiating each term, we get:

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

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

Rearrange to isolate $\frac{dy}{dx}$:

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

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

Step 4: Substitute $x = 1$

To find $\frac{dy}{dx}$ at $x = 1$, we need to determine the corresponding value of $y$ when $x = 1$.

Substitute $x = 1$ into the original equation:

$3(1)^2 + 2(1)y + y^2 = 2$

$3 + 2y + y^2 = 2$

$y^2 + 2y + 1 = 0$

$(y + 1)^2 = 0$

Thus, $y = -1$ when $x = 1$.

Step 5: Substitute $x = 1$ and $y = -1$ into $\frac{dy}{dx}$

$\frac{dy}{dx} = \frac{-6(1) - 2(-1)}{2(1) + 2(-1)}$

$= \frac{-6 + 2}{2 - 2}$

$= \frac{-4}{0}$

This expression is undefined due to division by zero, which suggests a vertical tangent line at $x = 1$, $y = -1$. Therefore, $\frac{dy}{dx}$ is undefined at this point.

Would you like a deeper explanation of any specific part, or do you have any questions?

Here are 5 additional related questions for further practice:

1. How does implicit differentiation help in solving for $\frac{dy}{dx}$ in terms of $x$ and $y$?
2. What does it mean if $\frac{dy}{dx}$ is undefined at a certain point?
3. Can you find the value of $\frac{dy}{dx}$ for other values of $x$ in similar problems?
4. How would the process change if the original equation had a higher degree for $x$ or $y$?
5. How can you interpret a vertical tangent line in the context of graphing?

Tip: Always double-check by substituting values back into the differentiated equation to confirm whether the answer aligns with expected behavior, such as undefined slopes for vertical tangents.