I have shown that $f(x) = 1-2x^2$ on [-1,1] has an invariant measure equivalent to lebesgue measure via the change of coordinates $h(x) = \sin\pi x/2$. (I.e. $g(x) = h^{-1}( f( h(x)))= 1 -2|x|$ has leb measure as an invariant measure so $m(A) = leb(h^{-1}(A))$ is the measure I want.
Possibly using the above, I want to prove $\lim_{n\to \infty} \frac{1}{n} \log|(f^n)'|$ converges a.e.
I think writing the limit as a telescoping sum and applying birkhoff may work but I'm not completely sure. Otherwise I'm told it is possible to compute the limit explicitly for g.
First, you can show that $(f,m)$ is measure equivalent to a 1-Bernoulli shift (call T the shift). In order to show this, you can consider $S:\{0,1\}^{\infty} \rightarrow[0,1]$ defined as $S = h \circ J$ where $J$ is defined as $J:\{0,1\}^{\infty} \rightarrow[0,1]$ $$ \left(x_0, x_1, \ldots\right) \longmapsto-1+\sum_{i=0}^{\infty} 2^{-i} P_{0 i} $$ where $P_{j i}=x_j+\ldots+x_i \bmod 2 \quad$ for $j \leqslant i.$ Note that $J$ makes the 1-side shift equivalent to the lebesgue measure on $[-1,1]$. Overall, you can check that for each sequence $x \in \{0,1\}^{\infty}$ we have $f \circ S = S \circ T.$
Once this is established, we can proceed. Take $x, y \in\{0,1\}^\infty$ s.t. $Jx-Jy=2^{-k}$; then $$ J \circ T^n x-J \circ T^n y=2^{-k+n} $$ (check this!). So we have $$ \begin{aligned} & \frac{f^n(S x)-f^n(S y)}{S_x-S_y} \\ & =\frac{S \circ T^n x-S \circ T^n y}{Sx-Sy}= \\ & =\frac{S \circ T^n x-S \circ T^n y}{Sx-Sy} \frac{Jx-J y}{J \circ T^n x-J \circ T^n y} \cdot 2^n\\ & \stackrel{k \rightarrow+\infty}{\longrightarrow} \frac{h^{\prime}\left(S T^n x\right)}{h^{\prime}(Sx)} 2^n=\frac{\cos \frac{\pi}{2} S T^n x}{\cos \frac{\pi}{2} Sx} 2^n = \frac{\cos \frac{\pi}{2} f^n y}{\cos \frac{\pi}{2} y} 2^n \end{aligned} $$
with $y = Sx$. Finally, we have
$$ \log \cos{ \frac{\pi}{2} f^n} \stackrel{L_2}{\longrightarrow} \int_{-1}^1 \log \cos\left( \frac{\pi}{2} y \right) m(\text{d}y)$$
because $(f,m)$ is exact (hence strongly mixing); this implies $\frac{1}{n}\log \cos{ \frac{\pi}{2} f^n}\stackrel{a.s.}{\longrightarrow} 0.$
Hence,
$$\frac{1}{n} \log \left|(f^n)'(y)\right| = \log 2 +\frac{1}{n} \log \frac{\cos \frac{\pi}{2} f^n (y)}{\cos \frac{\pi}{2} y} \stackrel{a.s.}{\longrightarrow}\log 2 $$