If $f(x)=f(1-x)$, then $f'(x)=-f'(1-x)$ If $f(x)=f(1-x)$, then
$$\frac{df(x)}{dx}=\frac{df(1-x)}{dx}=\frac{df(1-x)}{d(1-x)}\frac{d(1-x)}{dx}$$
and since $\frac{d(1-x)}{dx}=-1\implies d(1-x)=-dx$,
$$\frac{df(x)}{dx}=\frac{df(1-x)}{-dx}\cdot (-1)=\frac{df(1-x)}{dx}\ne -\frac{df(1-x)}{dx},$$
where's the mistake?

$f'(1-x)$ is $\left.\dfrac{df(t)}{dt}\right|_{t = 1-x} = \dfrac{df(1-x)}{d(1-x)}$, not $\dfrac{df(1-x)}{dx}$.

e.g. for $f(x) = x(x-1) =  x^2 -x$,
$$\begin{align*}
f(1-x) &= (1-x)^2 - (1-x)\\
&= x^2 - 2x + 1 - 1 + x\\
&= f(x)\\
f'(x) &= 2x - 1\\
f'(1-x) &= 2(1-x) - 1\\
&= 1 - 2x\\
&= -f'(x)\\
\frac{df(1-x)}{dx} &= \frac{d(x^2 - x)}{dx}\\
&= 2x - 1\\
&= f(x)
\end{align*}$$
