$$\left|\int_0^1f^3(x)dx - f^2(0)\int_0^1f(x)dx\right| \leq \max_{0 \leq x \leq1} |f'(x)| \left(\int_0^1f(x)dx\right)^2$$
here the exponents means the exponential and not the composition.
My attempt was very humble:
Mean value theorem:
$\int_0^1 f^3(x)dx = f^2(c) \int_0^1 f(x)dx$
this would simplify to:
$|f^2(c) - f^2(0)| \leq \max_{0 \leq x \leq 1} |f'(x)| \int_0^1 f(x)dx$
but this is hard to work on with
If $f$ is non-negative, then \begin{align*} \left|\int_0^1(f(x)^2-f(0)^2)f(x)\,dx\right|&= \left|\int_0^1\left(2\int_0^xf(u)f'(u)\,du\right)f(x)\,dx\right| \\ &\leq\int_0^1\left|2\int_0^xf(u)f'(u)\,du\right||f(x)|\,dx \\ &\leq\int_0^1\left(2\int_0^x|f(u)f'(u)|\,du\right)|f(x)|\,dx \\ &\leq\max_{x\in[0,1)}|f'(x)|\left(2\int_0^1\int_0^xf(u)f(x)\,du\,dx\right)\,. \\ \end{align*} Since $f$ is non-negative, by Fubini's theorem \begin{align*} \int_0^1\int_0^xf(u)f(x)\,du\,dx=\int_0^1\int_u^1f(u)f(x)\,dx\,du\,. \end{align*} Hence \begin{align*} 2\int_0^1\int_0^xf(u)f(x)\,du\,dx&=\int_0^1\int_0^xf(u)f(x)\,du\,dx+\int_0^1\int_u^1f(u)f(x)\,dx\,du \\ &=\int_0^1\int_0^xf(u)f(x)\,du\,dx+\int_0^1\int_x^1f(x)f(u)\,du\,dx \\ &=\int_0^1\int_0^1f(u)f(x)\,du\,dx \\ &=\left(\int_0^1f(x)\,dx\right)^2\,. \end{align*}