We are given the equation $xy = 1$, and we are asked to find the value of $y' + \frac{dy}{dx}$.

First, let's implicitly differentiate both sides of the equation $xy = 1$ with respect to $x$.

Step 1: Differentiate both sides

Using the product rule for differentiation: $\frac{d}{dx}(xy) = \frac{d}{dx}(1)$

The product rule says $\frac{d}{dx}(xy) = x \frac{dy}{dx} + y \frac{dx}{dx} = x \frac{dy}{dx} + y$, and the derivative of the constant $1$ is zero: $x \frac{dy}{dx} + y = 0$

Step 2: Solve for $\frac{dy}{dx}$

We can rearrange this equation to solve for $\frac{dy}{dx}$: $x \frac{dy}{dx} = -y$ $\frac{dy}{dx} = -\frac{y}{x}$

Step 3: Substitute into the original expression

We are asked to find $y' + \frac{dy}{dx}$, which is the same as $\frac{dy}{dx} + \frac{dy}{dx}$. From the differentiation, we know $y' = \frac{dy}{dx}$, so: $y' + \frac{dy}{dx} = \frac{dy}{dx} + \frac{dy}{dx} = 2 \cdot \frac{dy}{dx}$

Now, substitute $\frac{dy}{dx} = -\frac{y}{x}$ into this expression: $2 \cdot \frac{dy}{dx} = 2 \cdot \left(-\frac{y}{x}\right) = -\frac{2y}{x}$

Step 4: Conclusion

From the original equation $xy = 1$, we know that $y = \frac{1}{x}$. Substituting this into the expression: $-\frac{2y}{x} = -\frac{2 \cdot \frac{1}{x}}{x} = -\frac{2}{x^2}$

Thus, the expression for $y' + \frac{dy}{dx}$ simplifies to $-1$.

Therefore, the answer is: $\boxed{-1}$

Would you like more details on this, or do you have any other questions?

Here are some related questions to think about:

1. How do you apply the product rule in implicit differentiation?
2. What other ways could the equation $xy = 1$ be solved using derivatives?
3. What happens to the derivative when different functions of $x$ and $y$ are multiplied?
4. Can implicit differentiation be applied to more complex equations like trigonometric functions?
5. What role does the constant term play in implicit differentiation?

Tip: Always verify the basic identity or equation provided before applying differentiation techniques.