Let $f:\mathbb R\to\mathbb R$ be differentiable $t:\mathbb R\to(0,\infty)$ be a "step size" function and $$\tau:\mathbb R\to\mathbb R\;,\;\;\;x\mapsto x-t(x)f'(x).$$ In the steepest descent method, we iterate $\tau$ to find a local minimum of $f$. Let's consider$^1$ $f(x)=x^2(x-2)^2$ so that $f'(x)=4x(x-1)(x-2)$.
Note that the fixed points of $\tau$ are precisely the critical points of $f$. In our case, $x_0:=0$, $x_1:=1$ and $x_2:=2$.
How can we determine the stable and unstable manifold of $x_1$?
It's clear to me that $x_1$ is unstable$^2$, since obviously there is a $\varepsilon>0$ with $f'(x)>0$ for all $x\in(x-\varepsilon,x)$ and $f'(x)<0$ for all $x\in(x,x+\varepsilon)$ so that the iteration "is moving away" from $x_1$ in different directions depending on whether we start close to the left or to the right of $x_1$.
$^1$ $f$: 
$f'$:
$^2$ Recall that a set $A$ is stable if for every neighborhood $N$ of $A$, there is a neighborhood $U$ of $A$ such that $\bigcup_{n\in\mathbb N_0}\tau^n(U)\subseteq N$. A point $x$ is stable if $\{x\}$ is stable.