outline:
That would be trivial if $\psi$ had a smooth inverse. Since it is only a submersion this is not true, but you can make up for it as follows:
Since $\psi$ is a submersion, $\psi^{-1}(p)$ is a smooth submanifold of $M$ for each $p$ in the image of $\psi$. Now you need to know (or show) that there is locally, for $x\in \psi^{-1}(p)$ a manifold $C $ which is transversal to $\psi^{-1}(p)$, such that a neighbourhood $U$ of $x$ is diffeomorphic to a product $(U\cap \psi^{-1}(p))\times C$. (If you know about normal bundles and exponential maps this should be clear, otherwise you'll have to skim through what you already learned)
Now if you restrict $\psi$ to $C$ (more precisely: a copy of $C$ in $M$) it will be a diffeomorphism (why?), so that you can find a local inverse. That will help you to show that $\phi$ is smooth if $\phi\circ \psi$ is.
Edit in Response to a commment:
You need to know that (and this follows from the implicit function theorem or the inverse function theorem), if $D:=\psi^{-1}(p)$ is a smooth $k$- dimensional submanifold of $M$ (with $M$ having dimension $m= k+l$, say), then for each $x$ in $M$ there is a neighbourhood $U$ and a diffeomorphism $\xi: \mathbb{R}^k \times \mathbb{R}^l \rightarrow M\cap U$ such that $\xi(0) = x$ and
$$D\cap U = \xi (\mathbb{R}^k \times {0})$$
(i.e. you can straighten out the manifold $D$ locally in a chart).
The $C$ I've been talking about is then just $ \xi (\{0\} \times V^l$ for some sufficiently small neighbourhood $V^l \subset \mathbb{R}^l$ of $0$.
