Differentiable functions $f'(x)=f(-x)^4f(x)$ Find all differentiable functions $f\colon \mathbb{R}\to\mathbb{R}$ with $f(0)=1$ such that $f'(x)=f(-x)^4f(x)$, for all $x \in \mathbb{R}$.

The defining relation $f'(x)=f^4(-x)f(x)$ implies that $f$ is continuously differentiable (in fact it's even $C^\infty$).
Setting $-x$ in the relation $f'(x)=f^4(-x)f(x)$, one gets $\forall x, f'(-x) = f^4(x)f(-x)$.
Multiplying $f'(x)=f^4(-x)f(x)$ by $f^3(x)$ yields $$\forall x, f'(x)f^3(x) = f^4(-x)f^4(x)=[f^4(x)f(-x)]f^3(-x)=f'(-x)f^3(-x)$$
that is to say $x\mapsto f'(x)f^3(x) $ is even.
Therefore, $\displaystyle \int_{-x}^xf'(t)f^3(t) dt = 2\int_{0}^xf'(t)f^3(t) dt$, and since an antiderivative of $f'f^3$ is $\displaystyle \frac{f^4}4$, this implies $$\forall x, f^4(x)+f^4(-x)=2$$
Replacing $f^4(-x)$ in the defining relation, one gets $$\forall x, f'(x)=2f(x)-f^5(x)$$
By Picard–Lindelöf theorem, this non-linear differential equation (with initial condition $f(0)=1$) has a unique global solution. Using a computer or other means, one finds that  $$x\mapsto \frac{\sqrt[4]{2} e^{2 x}}{\sqrt[4]{e^{8 x}+1}}$$ is a solution of the differential equation, thus it must be the only one.
Conversely, it's easily checked that $x\mapsto \frac{\sqrt[4]{2} e^{2 x}}{\sqrt[4]{e^{8 x}+1}}$ is indeed a solution to the original functional equation.
