Fixed point versus point with slope $0$ When solving a first order differential equation
$$\frac{dy}{dx} = f(y)$$
for the fixed points (or steady states), we set the differential equation to $0$ and solve for the values of $y$ that are fixed points. Why does this guarantee that we will obtain fixed points? Why can't we obtain a point at which the slope is $0$ but is not a fixed point?
 A: A fixed point $y_0$ is a constant solution
$y(t) = y_0$ to the differential equation
$$\tag{1} y' = f(y).$$
First, if $y_0$ is a fixed point, then the constant function $y(t) = y_0$ has derivative $y'(t) = 0$. Using (1), we have
$$ f(y_0) = f(y(t)) = y'(t) = 0,$$
thus $f(y_0) =0$. On the other hand, if $y_0$ is a zero of $f$, then the constant functions $y(t) = y_0$ satisfies (1) and thus is a fixed point.
The following remark might be the reason why you are confused: there might be nonzero constant solutions passing through the zero of $f(y_0)$: the standard example is the ODE
$$y' = 3y^{2/3},$$
where $y_0 = 0$ is the zero of $f(y) = 3y^{2/3}$, however there is a non-constant solution $$ y(t) = t^3$$
passing through $y_0 = 0$ and is not a fixed point. That is, there is a solution passing through the zero of $f$ with zero slope. Your concern is real.
But this does not invalidate the claim that the any zero of $f$ contribute to a fixed point. The (more worrying) issue is that there are non-uniqueness of solutions to ODE. This could happen when $f$ is not continuously differentiable (note that in our example, $(y^{2/3})' = y^{-1/3}$ is not continuous at $y=0$).
When $f$ is continuously differentiable, the Picard-Lindelof theorem says that there is an unique solution to (1) with a given initial condition $y(t_0) = y_0$. Since we have checked that the constant function $y(t) = y_0$ is a solution to (1) with that given initial condition, the theorem implies that the constant solutions are the only possible solution.
