function such that $f(x\cdot t)=f(x)g(t)$ Let $E$ be the set of tuples of continuous functions $f,g:\mathbb{R}^*_+\rightarrow\mathbb{R}$ s.t. $f,g$ are never $0$ and $\forall x,t>0,f(x\cdot t)=f(x)g(t)$.
I need to show that $f,g\in\mathcal{C}^\infty$ and find $E$.

I found that $f=f(1)\cdot g$:
$f(x)=f(x)g(1)\Rightarrow f(xt)=f(x)g(1)g(t)\Rightarrow g(1)=1$
$f(x)=f(1)g(x)\Rightarrow f(xt)=f(1)g(x)g(t)\Rightarrow f(x)=f(1)g(x)$)
However, that does not give me the expected result.
 A: As you noticed, $g(1) = 1$ and $f(x) = f(1)g(x)$. Since $f(1) \neq 0$, we obtain $g(xt) = g(x) g(t)$. 
Now by induction, we see that $g(t^n) = g(t)^n$ and consequently $g(t^q) = g(t)^q$ for each $q \in \mathbb{Q}$. By continuity of $g$, $g(t^r) = g(t)^r$ for each $r \in \mathbb{R}$ and this implies that $g(t) = t^\alpha$ for some $\alpha \in \mathbb{R}$ (one way to see this is to write $g(t) = g(2^{\log_2 t}) = g(2)^{\log_2 t}= 2^{\log_2 g(2) \log_2 t} = t^{\log_2 g(2)}$).
Therefore, $g \in C^{\infty}$ and, since $f(t) = f(1) t^\alpha$, $f \in C^\infty$.
Conversely, given $\alpha \in \mathbb{R}$ and $A \neq 0$, we may set $g(t) = t^\alpha$ and $f(t) = At^\alpha$ to obtain $f(xt) = A(xt)^\alpha = f(x)g(t)$.
Thus, $$E = \{(f(t) = At^\alpha, g(t) = t^\alpha) | \alpha, A \in \mathbb{R}, A \neq 0\}.$$
A: If $g$ is zero at $t_0\in(0,\infty)$, then for all $x>0$ we have $f(x)=f(x/t_0\cdot t_0)=f(x/t_0)g(t_0)=0$, so that $f$ is identically zero. In that case, the equation does not tell you anything, and there is no reason for $g$ to be even once differentiable.
Similarly, if $f$ is identically zero, we cannot say anything about $g$ and if it vanishes ar $t_0$ and not at $t_1$, then $0=f(t_0)=f(t_1\cdot t_0/t_1)=f(t_1)g(t_0/t_1)$, so that $g$ vanishes at $t_0/t_1$ and we are backin the first case, where we know that $f$ is identically zero, against the hypothesis.
We thus see that either you suppose that botyh $f$ and $g$ are never zero or the conclusion you want is false.
A: I can't show differentiability, but if you assume it, then you can write
$$\frac{f(xt)-f(x)}{h} = \frac{f(x)}{x} \left( \frac{g(t)-1}{h/x} \right),$$
let $t=1+h/x$ and use the fact $g(1)=1$, and finally take the limit as $h \to 0$ to get
$$f'(x) = \frac{f(x)}{x} g'(1).$$
It's a hop, skip, and a jump to getting the solution $f(x) = A x^{g'(1)}$. Then since $f(x) = g(x)f(1)$, it is easy to verify this works.
