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We are tasked with using implicit differentiation to find the equation of the tangent line to the given curve at the point \((0, 0.5)\).

### The equation is:

\[
x^2 + y^2 = \left(3x^2 + 2y^2 - x\right)^2
\]

The given point on the curve is \((0, 0.5)\).

### Step 1: Implicit Differentiation
To find the slope of the tangent line, we need to implicitly differentiate both sides of the equation with respect to \(x\).

- Differentiate \(x^2 + y^2\):
\[
\frac{d}{dx}(x^2 + y^2) = 2x + 2y\frac{dy}{dx}
\]
- Differentiate the right-hand side \(\left(3x^2 + 2y^2 - x\right)^2\) using the chain rule:
\[
\frac{d}{dx}\left[\left(3x^2 + 2y^2 - x\right)^2\right] = 2\left(3x^2 + 2y^2 - x\right) \cdot \frac{d}{dx}(3x^2 + 2y^2 - x)
\]
Now, differentiate \(3x^2 + 2y^2 - x\):
\[
\frac{d}{dx}(3x^2 + 2y^2 - x) = 6x + 4y\frac{dy}{dx} - 1
\]

### Step 2: Substitute and Simplify
Now, substitute the derivatives into the equation:

\[
2x + 2y\frac{dy}{dx} = 2(3x^2 + 2y^2 - x) \cdot (6x + 4y\frac{dy}{dx} - 1)
\]

### Step 3: Evaluate at Point \((0, 0.5)\)
Substitute \(x = 0\) and \(y = 0.5\) into the differentiated equation:

- For the left side:
\[
2(0) + 2(0.5)\frac{dy}{dx} = 1\frac{dy}{dx}
\]
- For the right side:
\[
2(3(0)^2 + 2(0.5)^2 - 0) \cdot (6(0) + 4(0.5)\frac{dy}{dx} - 1)
\]
\[
= 2(2 \cdot 0.25) \cdot (2\frac{dy}{dx} - 1) = 1 \cdot (2\frac{dy}{dx} - 1)
\]

The full equation at \((0, 0.5)\) is:
\[
1\frac{dy}{dx} = 1(2\frac{dy}{dx} - 1)
\]
Simplify:
\[
\frac{dy}{dx} = 2\frac{dy}{dx} - 1
\]
\[
0 = \frac{dy}{dx} - 1
\]
Thus, \(\frac{dy}{dx} = 1\).

### Step 4: Equation of the Tangent Line
The slope of the tangent line at \((0, 0.5)\) is \(m = 1\). The equation of the tangent line can be written using the point-slope form:

\[
y - 0.5 = 1(x - 0)
\]

So, the equation of the tangent line is:

\[
y = x + 0.5
\]

### Final Answer:
\[
y = x + 0.5
\]

Let me know if you would like further clarifications or details!

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Here are 5 related questions:
1. How does implicit differentiation differ from explicit differentiation?
2. What are the key steps in using the chain rule during implicit differentiation?
3. How can we interpret the meaning of the slope of the tangent line geometrically?
4. How would the process change if the curve were parametrized?
5. What are some common pitfalls when applying implicit differentiation?

**Tip:** Always ensure that after applying implicit differentiation, you substitute the values of the given point correctly to avoid calculation errors.

Use implicit differentiation to find an equation of the tangent line to the curve at the given point: x² + y² = (3x² + 2y² - x)² at (0, 0.5).

Find the equation of the tangent line to the curve x² + y² = (3x² + 2y² - x)² at the point (0, 0.5) using implicit differentiation. The solution involves key calculus concepts like the chain rule and tangent line formula. Ideal for undergraduate students.

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