I'd like to find a function such that $f(x+y) = f(x)^2f(y)^2.$ Yes I realize initial conditions should be specified, but I don't have that luxury, I'm just looking for any function at all and then set the constant in the family of solutions later.
Here's my approach:
First, differentiate by $x$ to obtain
$f'(x+y) = 2f(x)f'(x)f(y)^2.$
Next I set $x = 0$ to find
$f'(y) = 2f(0)f'(0)f(y)^2.$
This is now an ODE which I solve for $f(y) = \frac{2f(0)f'(0)}{c_1 - x}.$
Now the problem is that $f(y+x) = \frac{2f(0)f'(0)}{c_1 - y - x} \neq \frac{2f(0)f'(0)}{c_1 - y} \cdot \frac{2f(0)f'(0)}{c_1-x}.$
Where did this technique go wrong?
It doesn't work because you can't assume differentiability, and also can't assume $f(y) \ne 0$.
Instead, if $f(y)=0$ for some $y$ then $f(x)=f^2(y)f^2(x-y)=0$ i.e. the trivial solution $f \equiv 0$.
Otherwise, it means $f(x) \ne 0$ for all $x$, then $f(0+0)=f^2(0)f^2(0) \implies f(0) = 1$, and it follows that $f(x+0)=f^2(x)f^2(0) \implies f(x) = 1$ i.e. the other trivial solution $f \equiv 1$.