Find 3 fixed points of function with 2 arguments

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I'm looking for a way to determine the fixed points of a function with 2 arguments. The specific function is the following:

$F:\bigl( \begin{smallmatrix} x \\ y \end{smallmatrix} \bigr) \mapsto \bigl( \begin{smallmatrix} 3*x*(1-x) - (2*x*y)/(1+x) \\ (7*x*y)/(2*(1+x)) \end{smallmatrix} \bigr)$

It's given that this function has 3 fixed points. But I have no clue on how to start finding these fixed points. I found a lot of information on fixed points for functions with only one argument that say that you just have to find the roots of the expression $f(x) - x = 0$, but nothing for a function of this form.

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$x$ gets mapped to $3x(1-x) - \frac{2xy}{1+x}$

$y$ gets mapped to $\frac{7xy}{2(1+x)}$

To find the fixed points, we'd like $f(x) - x = 0$ and $f(y) - y = 0$ simultaneously$^1$.

$$3x(1-x) - \frac{2xy}{1+x} - x = 0$$

$$\frac{7xy}{2(1+x)} - y = 0$$

This gives us the fixed points of $(0,0),(\frac{2}{3},0),(\frac{2}{5},\frac{14}{25})$


$^1$In actuality we want $F\bigl( \begin{smallmatrix} x \\ y \end{smallmatrix} \bigr) - \bigl( \begin{smallmatrix} x \\ y \end{smallmatrix} \bigr) = 0$, but this just becomes the two simultaneous equations I listed above. Notice how this is exactly the same as $F(x) - x = 0$, except here our "$x$" is a vector and not a scalar. It's the same concept.