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To find \( y' \) for the given equation \( x^2y - 3x + 2y^2 = 0 \) at the point \( (0,1) \), we'll differentiate the equation implicitly with respect to \( x \).

### Step-by-Step Solution:

1. **Given Equation:**
   \[
   x^2y - 3x + 2y^2 = 0
   \]

2. **Differentiate Implicitly with respect to \( x \):**
   Using the product rule for \( x^2y \) and the chain rule for \( y^2 \):
   \[
   \frac{d}{dx}(x^2y) - \frac{d}{dx}(3x) + \frac{d}{dx}(2y^2) = 0
   \]

   The derivatives are:
   - \( \frac{d}{dx}(x^2y) = 2xy + x^2\frac{dy}{dx} = 2xy + x^2y' \),
   - \( \frac{d}{dx}(-3x) = -3 \),
   - \( \frac{d}{dx}(2y^2) = 4y\frac{dy}{dx} = 4yy' \).

   Substituting these, we get:
   \[
   2xy + x^2y' - 3 + 4yy' = 0
   \]

3. **Simplify and Solve for \( y' \):**
   Combine all terms involving \( y' \):
   \[
   x^2y' + 4yy' = 3 - 2xy
   \]

   Factor out \( y' \):
   \[
   y'(x^2 + 4y) = 3 - 2xy
   \]

   Solve for \( y' \):
   \[
   y' = \frac{3 - 2xy}{x^2 + 4y}
   \]

4. **Substitute \( (x, y) = (0, 1) \):**
   At \( (0,1) \), substitute \( x = 0 \) and \( y = 1 \):
   \[
   y' = \frac{3 - 2(0)(1)}{0^2 + 4(1)}
   \]

   Simplify:
   \[
   y' = \frac{3}{4}
   \]

### Final Answer:
\[
y' = \frac{3}{4}
\]

Would you like me to explain any step in more detail?

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### Related Questions:
1. What is implicit differentiation, and why is it useful?
2. Can this method be applied to equations without explicit \( y \) terms?
3. What would happen if the point given were not on the curve?
4. How does the product rule work for terms like \( x^2y \)?
5. What is the geometrical interpretation of \( y' \) at a specific point?

### Tip:
When solving for \( y' \) in implicit differentiation, always isolate the terms with \( y' \) carefully to avoid errors in factoring.

Find y' for the equation x^2y - 3x + 2y^2 = 0 at the point (0, 1).

Learn how to find the derivative y' using implicit differentiation for the equation x^2y - 3x + 2y^2 = 0 at the point (0, 1). Step-by-step solution using product and chain rules. Suitable for high school and college calculus students.

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