Consider f(x,y) = (-3y+x^2, -2x). Find all fixed points of f

For fixed points, we want to solve f(x, y) = (x, y)
==> (-3y + x^2, -2x) = (x, y)
==> -3y + x^2 = x, and -2x = y.

Substitute the second equation into the first:
-3(-2x) + x^2 = x
==> x^2 + 5x = 0
==> x = 0 or -5
==> y = 0 or 10, respectively via y = -2x.

So, we have two fixed points: (x, y) = (0, 0), (-5, 10).

I hope this helps!

can put in any values in for (x,y) to get a value for f(x,y) but

if you mean to find the points where f(x,y)=(x,y) then you set up two equations
-3y+x^2=x
-2x=y now put this one into the first

-3(-2x)+x^2=x
6x+x^2=x
5x+x^2=0
x(x+5)=0
x=0 or x=-5 then putting back in to solve for y we get
(0,0) or (-5,10)

Dont know