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To find \( \frac{dx}{dt} \) given the equation
\[
y^2 + xy - 3x = -1,
\]
and knowing \( \frac{dy}{dt} = 2 \) when \( x = 1 \) and \( y = 1 \), we’ll use implicit differentiation with respect to \( t \).

### Step-by-Step Solution
1. **Differentiate both sides of the equation with respect to \( t \):**

   \[
   \frac{d}{dt}(y^2) + \frac{d}{dt}(xy) - \frac{d}{dt}(3x) = \frac{d}{dt}(-1).
   \]

2. **Apply the chain rule:**
   - For \( y^2 \), we have:
     \[
     \frac{d}{dt}(y^2) = 2y \cdot \frac{dy}{dt}.
     \]

   - For \( xy \), we use the product rule:
     \[
     \frac{d}{dt}(xy) = \frac{dx}{dt} \cdot y + x \cdot \frac{dy}{dt}.
     \]

   - For \( 3x \), we get:
     \[
     \frac{d}{dt}(3x) = 3 \cdot \frac{dx}{dt}.
     \]

   Putting it all together, the differentiated equation is:
   \[
   2y \cdot \frac{dy}{dt} + \left( \frac{dx}{dt} \cdot y + x \cdot \frac{dy}{dt} \right) - 3 \cdot \frac{dx}{dt} = 0.
   \]

3. **Substitute the known values \( x = 1 \), \( y = 1 \), and \( \frac{dy}{dt} = 2 \):**
   
   \[
   2(1)(2) + \left( \frac{dx}{dt} \cdot 1 + 1 \cdot 2 \right) - 3 \cdot \frac{dx}{dt} = 0.
   \]

   Simplifying each term:
   \[
   4 + \left( \frac{dx}{dt} + 2 \right) - 3 \cdot \frac{dx}{dt} = 0.
   \]

4. **Combine like terms:**
   
   \[
   4 + 2 - 2 \cdot \frac{dx}{dt} = 0.
   \]

   This simplifies further to:
   \[
   6 - 2 \cdot \frac{dx}{dt} = 0.
   \]

5. **Solve for \( \frac{dx}{dt} \):**
   
   \[
   2 \cdot \frac{dx}{dt} = 6 \Rightarrow \frac{dx}{dt} = 3.
   \]

### Final Answer
\[
\frac{dx}{dt} = 3.
\]

Would you like further details or have any questions?

---

Here are some follow-up questions for deeper understanding:

1. Why did we use implicit differentiation for this problem?
2. How does the product rule apply when differentiating \( xy \) with respect to \( t \)?
3. What are other applications of implicit differentiation in calculus?
4. How would the solution change if \( \frac{dy}{dt} \) had a different value?
5. What would happen if we needed to find \( \frac{d^2x}{dt^2} \) as well?

**Tip:** When differentiating implicitly, always remember to apply the chain rule carefully for each term.

Suppose that x = x(t) and y = y(t) are both functions of t. If y^2 + xy - 3x = -1, and dy/dt = 2 when x = 1 and y = 1, what is dx/dt?

Learn how to find dx/dt using implicit differentiation of the equation y^2 + xy - 3x = -1 with given values of x, y, and dy/dt. This solution demonstrates the use of the chain rule and product rule. Ideal for high school calculus students in Grades 11-12.

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