Even if $|f'(u)| = 1$, the following criteria still work:
The fixed point $u$ is locally stable if there exists a neighborhood $U$ of $u$ such that $|f(x)-u| < |x-u|$ for all $x \in U$, $x \ne u$.
The fixed point $u$ is locally unstable if there exists a neighborhood $U$ of $u$ such that $|f(x)-u| > |x-u|$ for all $x \in U$, $x \ne u$.
Often, it helps to look at the graph of the function $f$ (here $f(x)=1+\log x$) near the fixed point:
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f near its fixed point">
If the graph of the function near the fixed point lies entirely within green area delimited by the line $y=x$ and its reflection through the fixed point, the fixed point is locally stable. If the graph lies entirely in the red area near the fixed point, it is locally unstable.
If, as in this example, the graph of $f$ lies in the green area on one side of the fixed point and in the red area on the other, things get more complicated. If, like here, $f$ is increasing near the fixed point, it will be locally stable on one side and unstable on the other, and is thus a saddle point. If $f$ is decreasing, you'll need to repeat the stability analysis for its second iterate $f^{(2)}$ (which will be increasing) and see what happens to it.
Finally, if the graph of $f$ lies on the line $y=x$, it has a continuum of neutrally stable fixed points, while if it lies on the reflected line (or, more generally, if the graph of $f^{(2)}$ lies on the $y=x$ line), it will have a single neutrally stable fixed point surrounded by neutral 2-cycles.
Of course, just looking at the graph is in general not enough: you have to prove that the graph doesn't cross over into another area even closer to the fixed point. But as long as $f$ is fairly well-behaved, looking at the graph at least gives you a good idea what the answer is going to be.
Since this is homework, I won't spoon-feed you the answers, but you should be able to figure them out using the criteria I gave above.