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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 \(x\).

If -y^2 + 2x^2 = -2, then find dy/dx at the point (1, -2) in simplest form.

Learn how to find the derivative dy/dx using implicit differentiation for the equation -y^2 + 2x^2 = -2 at the point (1, -2). This calculus problem covers implicit differentiation concepts, including the chain rule and derivative formulas. Suitable for students in Grades 11-12.

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