My attempt is based on giving an explicit function between both sets but through compositions, which is up to what I have learned so far.
First, let us consider the map $ f: (0,1) \times (0,1) \to (0,1) $, defined for two real numbers $ x, y \in (0,1) $, with their respective digits $ 0, x_1x_2 \ldots $ and $ 0,y_1y_2 \ldots $. The map is given by $$ f (x, y) = 0, x_1y_1x_2y_2 \ldots $$ What needs to prove is that this function is bijective, I don't understand how to do that, some help for that part.
Now consider the map $ \bar {f} :( 0,1) \to \mathbb R $ given by $ \bar {f} (x) = \ln \left (\dfrac{1}{x}-1 \right) $, this it is clearly bijective.
Let $ \bar {g}: \mathbb R \to (0,1) $ given by $ \bar {g} (x) = 1 / (1 + e ^ x) $, clearly bijective.
In this way we define the function $ \bar {G}: \mathbb R \times \mathbb R \to (0,1) \times (0,1) $ given by $ \bar {G} (x, y) = (\bar {g} (x), \bar {g} (y)) $, also clearly bijective.
Therefore we will have the following $$ \mathbb R \times \mathbb R \overset {\bar {G}} {\to} (0,1) \times (0,1) \overset {f} {\to } (0,1) \overset {\bar {f}} {\to} \mathbb R $$ The searched function is $\phi:\mathbb R\times \mathbb R\to \mathbb R$ given by $\phi=\bar{f}\circ f\circ \bar{G}$.
But since functions of two variables are composed with functions of a single variable, they could help me in that part as well. Thank you very much for reading my answer.
