To find the derivative of the implicit function $xy + x = 3y$ with respect to $x$, we'll use implicit differentiation. Here’s the step-by-step process:

Given the equation: $xy + x = 3y$

1. Differentiate both sides of the equation with respect to $x$:

$\frac{d}{dx}(xy) + \frac{d}{dx}(x) = \frac{d}{dx}(3y)$

2. Apply the product rule to the term $xy$:

$\frac{d}{dx}(xy) = y \cdot \frac{d}{dx}(x) + x \cdot \frac{d}{dx}(y) = y + x \frac{dy}{dx}$

3. The derivative of $x$ with respect to $x$ is 1, and the derivative of $3y$ with respect to $x$ is $3 \frac{dy}{dx}$:

$y + x \frac{dy}{dx} + 1 = 3 \frac{dy}{dx}$

4. Now, isolate $\frac{dy}{dx}$:

$x \frac{dy}{dx} - 3 \frac{dy}{dx} = -y - 1$

$\frac{dy}{dx}(x - 3) = -y - 1$

5. Finally, solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{-y - 1}{x - 3}$

So, the derivative of $y$ with respect to $x$ is: $\frac{dy}{dx} = \frac{-y - 1}{x - 3}$

Would you like more details on any specific step or have any questions?

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Here are five related questions for further exploration:

1. How would you find the second derivative of $y$ with respect to $x$ for the same equation?
2. What is the geometric interpretation of the derivative $\frac{dy}{dx}$ in this context?
3. How does implicit differentiation differ from explicit differentiation?
4. Can you solve for $y$ explicitly in terms of $x$ from the given equation?
5. How would you find the points where the tangent line to the curve is horizontal?

Tip: When dealing with implicit differentiation, remember to apply the product rule carefully when differentiating terms like $xy$.