$\newcommand{\pa}{\partial}\newcommand{\diff}{\text{diff}}$Since $f$ is an increasing diffeomorphism, we have $f'>0$, so that
\begin{equation*}
f'=e^h
\end{equation*}
for some real-valued function $h$. Then $f''=e^h h'$ and hence $h''=(f''/f')'<0$, by one of the given conditions on $f$. So, the function $h$ is strictly concave. Also, the condition $f''(0)=0$ and the identity $f''=e^h h'$ imply $h'(0)=0$, so that the strictly concave function $h$ is (strictly) decreasing. Moreover, the conditions $f(0)=0$ and $f(1)=1$ imply
\begin{equation*}
1=\int_0^1 f'=\int_0^1 e^h<e^{h(0)},
\end{equation*}
since $h$ is decreasing. So,
\begin{equation*}
h(0)>0. \tag{10}\label{10}
\end{equation*}
Take now any $x\in(0,1)$. By the strict concavity of $h$, for all $t\in(x,1)$,
\begin{equation*}
h(t)<h_x(t):=h(0)+\frac tx\,[h(x)-h(0)].
\end{equation*}
Therefore and because $f(1)=1$, we have
\begin{equation*}
\begin{aligned}
f(x)&=1-\int_x^1 dt\, f'(t) \\
&=1-\int_x^1 dt\, e^{h(t)} \\
&>1-\int_x^1 dt\, e^{h_x(t)} \\
& =1-x e^{h(x)}\frac{1-e^{-(1/x-1)[h(0)-h(x)]}}{h(0)-h(x)} \\
& =1-x e^u R ,
\end{aligned}
\end{equation*}
where
\begin{equation*}
R:=R(v):=\frac{1-e^{-z(v-u)}}{v-u},\quad u:=h(x),\quad v:=h(0),\quad z:=1/x-1,
\end{equation*}
so that
\begin{equation*}
z>0\quad\text{and}\quad v>\max(0,u),
\end{equation*}
in view of \eqref{10} and because $h$ is decreasing.
Note that $(v-u)^2 e^{z (v-u)}R'(v)=1+z (v-u)-e^{z (v-u)}<0$, so that $R(v)$ is decreasing in $v$. Since $v>\max(0,u)$, we have $R(v)\le R(u+)=z$ if $u\ge0$ and $R(v)\le R(0)$ if $u<0$. So,
\begin{equation*}
f(x)>\left\{
\begin{alignedat}{2}
f_1(u)&:=\max[0,1-(1-x)e^u] &&\text{ if } u\ge0, \\
f_2(u)&:=1-xe^u\frac{1-e^{z u}}{-u} &&\text{ if } u<0.
\end{alignedat}
\right.
\end{equation*}
Let
\begin{equation*}
r:=\frac{1-f(x)^2}{f'(x)(1-x^2)}=\frac{1-f(x)^2}{e^u(1-x^2)}
\end{equation*}
and
\begin{equation*}
\diff:=1-f(x)^2-f'(x)(1-x^2)=1-f(x)^2-e^u(1-x^2).
\end{equation*}
We want to show that $r<1$ or, equivalently, $\diff<0$.
Consider first the case $u\ge0$. If now $(1-x)e^u\le1$, then $f_1(u)=1-(1-x)e^u$ and hence
\begin{equation*}
r<\frac{1-f_1(u)^2}{e^u(1-x^2)}=\frac{e^u (x-1)+2}{1+x}
\le\frac{ (x-1)+2}{1+x}=1,
\end{equation*}
so that $r<1$ indeed. If $(1-x)e^u>1$, then $f_1(u)=0$ and hence
\begin{equation*}
r<\frac{1-f_1(u)^2}{e^u(1-x^2)}=\frac{1}{e^u(1-x^2)}
<\frac{1}{1+x}<1,
\end{equation*}
so that we still have $r<1$.
Consider finally the case $u<0$. Here
\begin{equation*}
\diff<
H(x):=1-f_2(u)^2-u(1-x^2)
=1-(1-x^2) e^{w x}-\frac{(e^{w x}+w-e^w)^2}{w^2},
\end{equation*}
where $w:=u/x<0$. Let
\begin{equation*}
H_1(x):=\frac{-w}{e^{w x}}\,H'(x)=w (1-x) (w x+w+2)+2 e^{w x}-2 e^w.
\end{equation*}
Then $H_1'(x)=-2 w (1+w x-e^{w x})<0$. So, $H_1$ is decreasing in $x\in(0,1)$, to $H_1(1)=0$. So, $H_1>0$. So, $H(x)$ is increasing in $x\in(0,1)$, to $H(1)=0$. So, $H(x)<0$ and hence $\diff<0$. $\quad\Box$
Remark: Inequality
$$\frac{1-f(x)^2}{1-x^2}\le f'(x)$$
in the previous post does not make sense for $x=1$, but the slightly more general inequality
\begin{equation*}
1-f(x)^2\le f'(x)(1-x^2), \tag{20}\label{20}
\end{equation*}
whose strict version we have proved for $x\in(0,1)$, obviously holds for all $x\in[0,1]$, by continuity. In fact, \eqref{20} will still hold if the condition $(f''/f')'<0$ is relaxed to the condition that $f'$ is log concave.
