Show that there exists a unique $\psi:\mathbb{R}\to\mathbb{R}$ such that $\psi^3+e^x\psi=id$ holds. Furthermore, show that $\psi\in C^1(\mathbb{R},\mathbb{R})$ and compute $\psi'(0)$.
I'm trying to apply the Banach fixed-point theorem to $C^1$ as a subspace of all functions from $\mathbb{R}$ to $\mathbb{R}$, with a suitable metric, which would give us the first part of the exercise. However, I don't know what metric to choose so that the vector space of real functions is complete with respect to this norm and $C^1$ is closed as well (which we need in order to be able to apply the fixed point theorem).
For the second part: I have tried to compute $\psi'(0)$ using the identity $\psi^3+e^x\psi=id$ but the $\psi^3$ seems to be really problematic here.
Let $F(x, \psi)=\psi^3+e^x\psi-x$. We can calculate that $$\frac{\partial F}{\partial \psi}=3\psi^2+e^x>0$$ for each $(x, \psi)\in\mathbb{R}^2$. By implicit function theorem, the $\psi$ defined by $F(x,\psi)=0$ exists uniquely in $\mathbb{R}$ and its derivative $$\psi'=-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial \psi}}=-\frac{e^x\psi-1}{3\psi^2+e^x}$$ also exists everywhere. With $F(0, \psi(0))=\psi(0)^3+\psi(0)=0$ we get $\psi(0)=0$, and the problem is solved.