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The uploaded image shows a mathematical problem related to finding tangent lines to a given curve. Since the image isn't fully clear, I will try to extract the visible elements of the problem:

The equation provided seems to be:

\[
x^2 = y^3 - 2y + 1
\]

The goal appears to be finding the points where the tangent lines to this curve are horizontal.

### Steps to solve:

1. **Implicit Differentiation**: 
   To find horizontal tangents, we need to first differentiate the given curve implicitly with respect to \(x\) to find the slope \(\frac{dy}{dx}\).

2. **Set the Derivative Equal to Zero**: 
   For horizontal tangents, the slope of the tangent line (\(\frac{dy}{dx}\)) must be zero. We'll solve for \(y\) in this condition.

3. **Find the Corresponding \(x\)-Coordinates**: 
   After determining the \(y\)-coordinates, we substitute these values back into the original equation to find the corresponding \(x\)-coordinates.

Let me start solving it. I will perform implicit differentiation on the given equation:

\[
x^2 = y^3 - 2y + 1
\]

Differentiating both sides with respect to \(x\).The implicit differentiation of the equation \(x^2 = y^3 - 2y + 1\) results in:

\[
2x = 3y^2 \frac{dy}{dx} - 2 \frac{dy}{dx}
\]

Now, to find where the tangent lines are horizontal, we need to set \(\frac{dy}{dx} = 0\). For this, I will simplify the expression and solve for the values of \(y\) that make the derivative zero.The equation to find horizontal tangents simplifies to:

\[
3y^2 - 2 = 0
\]

Now, let's solve this equation for \(y\).The solutions for \(y\) where the tangent lines are horizontal are:

\[
y = \frac{-\sqrt{6}}{3} \quad \text{and} \quad y = \frac{\sqrt{6}}{3}
\]

Next, we'll substitute these \(y\)-values back into the original equation to find the corresponding \(x\)-coordinates. Let's compute those now.It seems like I can't do more advanced data analysis right now. Please try again later.

However, from the steps completed so far, you have the \(y\)-values where the tangent lines are horizontal: 
\[
y = \frac{-\sqrt{6}}{3} \quad \text{and} \quad y = \frac{\sqrt{6}}{3}
\]

To proceed, you can substitute these values back into the original equation \(x^2 = y^3 - 2y + 1\) to find the corresponding \(x\)-coordinates.

If you'd like more assistance or further clarification, feel free to ask!

Here are some related questions that might help deepen your understanding:

1. How do we interpret the geometric meaning of a horizontal tangent line?
2. What does implicit differentiation reveal about the relationship between \(x\) and \(y\)?
3. Can vertical tangents be found similarly? How would you approach that?
4. What role does symmetry play in finding the points of tangency?
5. How can the results be confirmed graphically?

**Tip:** When solving equations with square roots, always consider both the positive and negative roots to avoid missing solutions.

Find the points where the tangent lines to the curve are horizontal for x^2 = y^3 - 2y + 1.

Learn how to find the points where the tangent lines to the curve x^2 = y^3 - 2y + 1 are horizontal by using implicit differentiation and solving dy/dx = 0.

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