To solve the given problem, we start with the given equation:

$xy = 1$

We need to find the value of $y' + \frac{dy}{dx}$.

Step 1: Differentiate both sides with respect to $x$

Using the product rule, which states that if $u$ and $v$ are functions of $x$, then:

$\frac{d(uv)}{dx} = u \frac{dv}{dx} + v \frac{du}{dx},$

we differentiate $xy = 1$:

$\frac{d}{dx}(xy) = \frac{d}{dx}(1).$

Applying the product rule:

$x \frac{dy}{dx} + y \frac{dx}{dx} = 0.$

Simplify:

$x \frac{dy}{dx} + y = 0.$

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

Rearrange the equation:

$x \frac{dy}{dx} = -y.$

Divide by $x$:

$\frac{dy}{dx} = -\frac{y}{x}.$

Step 3: Substitute into $y' + \frac{dy}{dx}$

Since $y' = \frac{dy}{dx}$, the expression simplifies to:

$y' + \frac{dy}{dx} = \frac{dy}{dx} + \frac{dy}{dx} = 2 \frac{dy}{dx}.$

Substituting $\frac{dy}{dx} = -\frac{y}{x}$:

$2 \frac{dy}{dx} = 2 \left(-\frac{y}{x}\right) = -\frac{2y}{x}.$

Given $xy = 1$, we can solve for $y$:

$y = \frac{1}{x}.$

Substituting this value of $y$:

$-\frac{2y}{x} = -\frac{2 \cdot \frac{1}{x}}{x} = -\frac{2}{x^2}.$

Since the question specifically asks to find $y' + \frac{dy}{dx} = 0$, we equate the expressions above:

$y' + \frac{dy}{dx} = 0.$

Final Answer: (A)

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