Let's solve the problem:

We are given the equation:

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

and we need to find $\frac{dy}{dx}$ at the point $(1, -2)$.

Step 1: Differentiate implicitly with respect to $x$.

Implicit differentiation is used because $y$ is a function of $x$.

Starting with the equation: $- y^2 + 2x^2 = -2$

Differentiating each term with respect to $x$:

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

This gives:

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

Step 2: Solve for $\frac{dy}{dx}$.

Now isolate $\frac{dy}{dx}$:

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

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

Step 3: Plug in the point $(1, -2)$.

Substitute $x = 1$ and $y = -2$ into the expression:

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

Thus, $\frac{dy}{dx} = -1$ at the point $(1, -2)$.

The answer is $-1$.

Do you have any questions or would you like further details?

Here are 5 related questions to deepen your understanding:

1. How do you perform implicit differentiation for more complex functions?
2. What does it mean geometrically when $\frac{dy}{dx} = -1$?
3. How do you find the second derivative $\frac{d^2y}{dx^2}$ implicitly?
4. What are some common mistakes to avoid during implicit differentiation?
5. How can we confirm if a point lies on the original curve?

Tip: When doing implicit differentiation, always remember to apply the chain rule to terms involving $y$, since $y$ is a function of $x$.