To begin, if $xy = \frac{x}{y}$, we can conclude that $y\neq0$, otherwise $\frac{x}{y}$ is undefined.

Take: $xy = \frac{x}{y}$
Multiply both sides of the equation by $y$ to get: $xy^2 = x$
Subtract $x$ from both sides to get: $xy^2 - x = 0$
Factor to get: $x(y^2 - 1) = 0$
Factor again to get: $x(y - 1)(y + 1) = 0$
There are 3 POSSIBLE solutions to the above equation: $x = 0$, $y = 1$ and $y = -1$

Let's examine each possible case:

case i: $x = 0$
Substitute this into the original equation to get: $0 + y = (0)y = \frac{0}{y}$
For this equation to hold true, we need $y = 0$, but we already showed that $y$ cannot equal $0$
So, it can't be the case that $x = 0$

case ii: $y = 1$
Substitute this into the original equation to get: $x + 1 = x(1) = \frac{x}{1}$
Simplify: $x + 1 = x = x$
Since there are no solutions to the equation $x + 1 = x$, it can't be the case that $y = 1$

case iii: $y = -1$
Substitute this into the original equation to get: $x + (-1) = x(-1) = \frac{x}{-1}$
Simplify: $x - 1 = -x = -x$
Now take: $x - 1 = -x$
Add $x$ to both sides: $2x - 1 = 0$
Add $1$ to both sides: $2x = 1$
Solve: $x = 0.5$
So, $x = 0.5$ and $y = -1$ is the ONLY possible solution to the given equation.

Answer: B

Cheers,
Brent