There is the well-known characterization that random variables $X,Y$ are independent if and only if $F_{X,Y}(x,y) = F_X (x) \cdot F_Y (y)$. Due to a typo in a draft for an exam that I recently checked, this question was mis-stated as follows:
If $X,Y$ are real-valued random variables such that $F_{X \cdot Y} = F_X \cdot F_Y$, does it follow that $X,Y$ are stochastically independent?
Here, $F_{X \cdot Y}$ is the distribution function of the product $X \cdot Y$, i.e., $F_{X\cdot Y}(t) = \Bbb{P}(X\cdot Y \leq t)$. The assumption means more explicitly that $F_{X \cdot Y}(t) = F_X(t) \cdot F_Y (t)$ for all $t \in \Bbb{R}$.
I would guess that the answer to the question is "no". The trouble I have with proving this is that it is difficult for me to construct at all random variables $X,Y$ (independent or not) which satisfy $F_{X \cdot Y} = F_X \cdot F_Y$, except when $X$ or $Y$ are (almost surely) constant, in which case the claimed independence trivially holds.
One further observation: If $X,Y$ are indeed independent, then $$ F_X (t) \cdot F_Y(t) = \Bbb{P}(X \leq t) \cdot \Bbb{P}(Y \leq t) = \Bbb{P}(X \leq t \text{ and } Y \leq t) = \Bbb{P}(\max\{X,Y\} \leq t) = F_{\max \{X,Y\}}(t). $$ Thus, if the claim was true, we would have $F_{X \cdot Y} = F_{\max \{X,Y\}}$, meaning that $X \cdot Y \sim \max \{X,Y\}$. But even using this, I am unable to derive a contradiction.
Unless I made an arithmetic error, taking $X = Y$ with cdf $$F_X(t) = \begin{cases} 0,\ t \leqslant 1\\ \exp\left(\frac{-1}{\log t}\right) \end{cases}$$ gives a counterexample.
$$P(X \cdot Y \leqslant t) = P(X^2 \leqslant t) = P(X \leqslant \sqrt{t}) =\\ F_X(\sqrt{t})=\\ \exp\left(\frac{-1}{\log\sqrt{t}}\right)= \exp\left(\frac{-2}{\log t}\right) = \left(\exp\left(\frac{-1}{\log t}\right)\right)^2=\\ F_X(t)^2 = F_X(t)\cdot F_Y(t) = P(X \leqslant t) \cdot P(Y \leqslant t) $$
(to find this example I assumed $X = Y$, got equation $f(\sqrt{t}) = f(t)^2$ and asked wolfram)