Suppose $(M,d)$ is metric. I have proven that if $\psi\colon[0,\infty)\to[0,\infty)$ is non-decreasing, subadditive and satisfies $\psi(x)=0\iff x=0$ for $x\ge0$, then $$\rho(x,y)=\psi(d(x,y))$$ is a metric on $M$.
But I want to `tweak’ the restriction of this statement, I want to use differentiability of $\psi$. I found the following:
Suppose $(M,d)$ is metric and $\psi\colon[0,\infty)\to[0,\infty)$ is
differentiable with continuous non-increasing derivative $\psi'$ and
$\psi(0)=0$. Then $\psi$ is non-decreasing and subadditive.
I have also proven this statement, but is $\rho=\psi(d(x,y))$ a metric in this situation? And if not, what is a sufficient condition on the derivative $\psi'$ to turn $\rho$ into a metric on $M$?
You need to explicitly exclude $\psi'(0)=0$ (and thus due to the range of $\psi$ it follows $\psi'(0)\gt 0$) or alternatively require $\psi'$ strictly decreasing in any interval $(0,\epsilon)$. Otherwise, $\psi$ might satisfy the starting conditions of your second statement, but can violate $\psi(x)=0\iff x=0$, because if $\psi'$ can be zero on a positive interval starting at $0$, $\psi$ would be constant on that interval.
Then, from the starting conditions, $\psi' > 0$ on an interval near $0$ and thus $\psi$ strictly increasing, so $\neq 0$ on that interval. Now, from the given range of $\psi$ and $\psi'$ non-increasing it follows that $\psi' \ge 0$ everywhere – if it was not, then you can derive that $\psi$ must eventually become negative, which is a contradiction.
That in turn implies $\psi$ is strictly increasing near $0$ and (not necessarily strictly) increasing everywhere, and thus the remaining needed precondition of the first statement $\psi(x)=0\iff x=0$ follows.