Problem 3-1: Let $M$ and $N$ be smooth manifolds with or without boundary, and let $F:M\to N$ be a smooth map. Show that $dF_p:T_pM\to T_{F(p)}N$ is the zero map if and only if $F$ is constant on each component of $M$.
This is my attempt:
In local coordinates, we have $$dF_p\left(\frac{\partial}{\partial x^i}\Big|_p\right)=\frac{\partial F^j}{\partial x^i}(p)\frac{\partial}{\partial x^j}\Big|_{F(p)}\,.$$ Hence, $dF_p=0$ if and only if, for all $i$, $dF_p\left(\frac{\partial}{\partial x^i}|_p\right)=0$, if and only if $\frac{\partial F^j}{\partial x^i}(p)=0$ for all $i,j$, if and only if $F^j=$ constant for all $j$.
But after looking at this question I can see I am wrong and the solution is much more complicated.
Why is my answer wrong?
You seem to be jumping from the fact that all derivatives of $F$ at one point $p$ are $0$ to the conclusion that $F$ is constant. That is false, of course.
Even if you prove that all derivatives of $F$ at all points are $0$, then, when you start to work in local coordinates, you are working in with a chart $\psi\colon\mathbb R^n\longrightarrow M$. But then all that you shall have proved would be that the restriction of $F$ to $\psi(\mathbb R^n)$ is constant. You still will not have proved that the restriction of $F$ to each connected component of $M$ is constant.