Assume that $f:[0,1]\to [0,1]$ is an diffeomorphism so that $(f''(x)/f'(x))'<0$ and that $f''(0)=0$. It seems to me that $$\frac{1-f(x)^2}{1-x^2}\le f'(x)$$ on $[0,1]$. But no proof so far.
The answer posted below is correct. I however need $f$ to be increasing.
For $x\in[0,1]$, let
\begin{equation*}
f(x):=c\int_x^1 e^{-t^2}\,dt,
\end{equation*}
where $c:=1/\int_0^1 e^{-t^2}\,dt$.
Then $f\colon[0,1]\to [0,1]$ is a diffeomorphism such that $(f''(x)/f'(x))'<0$ for all $x\in[0,1]$ and $f''(0)=0$.
However, the inequality in question,
\begin{equation*}
L(x):=\frac{1-f(x)^2}{1-x^2}\le f'(x), \tag{1}\label{1}
\end{equation*}
fails to hold for any $x\in[0,1]$, since $f'<0$ and $L>0$ on $[0,1)$, whereas $L(1)$ is undefined.
