A fixed point of a function $f$ means that $f(x) = x$. In this instance, you want $f(x_{n-1}, y_{n-1}) = (x_n, y_n)$, so if $(x,y)$ is a fixed point, this means that you want $f(x,y) = (x,y)$, i.e.
$x = x + 0.004x + 0.002xy$
$y = y + 0.005y + 0.010xy - 0.001y^2$
So all you have to do is to solve for $x$ and $y$, the most trivial solution being $x = 0$ and $y = 0$, but there are more than one solution.
For the first equation, there are two solutions, either $x = 0$ or $0.004 + 0.002y = 0$
For the second equation, there are two solutions, either $y = 0$ or $0.005 + 0.010x - 0.001y = 0$.
All you have to do now is to go through all the possibilities of what $x$ and $y$ can be. $x = 0$ and $y = 0$ is the one most easily spotted, try to find the others.
A fixed point of a function $f$ means that $f(x) = x$. In this instance, you want $f(x_{n-1}, y_{n-1}) = (x_n, y_n)$, so if $(x,y)$ is a fixed point, this means that you want $f(x,y) = (x,y)$, i.e.
$x = x + 0.004x + 0.002xy$
$y = y + 0.005y + 0.010xy - 0.001y^2$
So all you have to do is to solve for $x$ and $y$, the most trivial solution being $x = 0$ and $y = 0$, but there are more than one solution.
To find all solutions, we solve the equations:
$0 = 0.004x + 0.002xy = x(0.004 + 0.002y)$
$0 = 0.005y + 0.010xy - 0.001y^2 = y(0.005 + 0.010x - 0.001y)$
For the first equation, there are two solutions, either $x = 0$ or $0.004 + 0.002y = 0$
For the second equation, there are two solutions, either $y = 0$ or $0.005 + 0.010x - 0.001y = 0$.
All you have to do now is to go through all the possibilities of what $x$ and $y$ can be. $x = 0$ and $y = 0$ is the one most easily spotted, try to find the others.