First of all, since $g(f(x)) = x^3$ and $x^3$ is injective, $f$ is injective. Indeed,
$f(x) = f(y) \implies g(f(x)) = g(f(y)) \iff x^3 = y^3 \iff x = y$. Plugging
$f(x)$ in $f(g(x)) = x^2$, we obtain $f(g(f(x))) = f(x)^2 \iff f(x^3) = f(x)^2$. If
$f, g: R \to R$, consider $x = -1, 0, 1$ (i.e., the roots of $x^3 = x$): $f(-1) = f(-1)^2$,
$f(0) = f(0)^2$ and $f(1) = f(1)^2$, so $\{f(-1), f(0), f(1)\} \subset \{0, 1\}$, which
contradicts the fact that $f$ is injective. So in this case there are no such
functions $f, g$.

If $f, g: S \to S$, consider $f(x^3) = (f(x))^2$ and $g(f(g(x))) = (g(x))^3 \iff$
$g(x^2) = (g(x))^3$. Applying these two identities $n$ times we obtain $f(x^{3^n}) =$
$(f(x))^{2^n}$ and $g(x^{2^n}) = (g(x))^{3^n}$. In particular, $f(2^{3^n}) = (f(2))^{2^n}$ and
$g(2^{2^n}) = (g(2))^{3^n}$. Extend these identities for all $n$ real. We have $x =$
$2^{3^n} \iff n = \log_3 2 \log_2 x$ and $x = 2^{2^n} \iff n = \log_2 \log_2 x$. This means
that $f(x) = a^{2^{\log_3 2 \log_2 x}}$ and $g(x) = b^{2^{\log_2 2 \log_2 x}}$. For the sake of simplicity, write all
equations in base $2$: $f(x) = 2^{2^{\log_3 2 \log_2 \log_2 x+k}}$ and $g(x) = 2^{2^{\log_2 3 \log_2 \log_2 x+\ell}}$.
Now substitute in the original equations:
$$\begin{align*} f(g(x)) &= f(2^{2^{\log_2 3 \log_2 \log_2 x + \ell}}) = 2^{2^{\log_3 2 \log_2 \log_2 2^{2^{\log_2 3 \log_2 \log_2 x + \ell}}}+ k} = 2^{2^{\log_3 2 \cdot (\log_2 3 \log_2 \log_2 x + \ell) + k}} \\ &= 2^{2^{\log_2 \log_2 x + \ell} \log_3 2 + k} = 2^{2^{\log_2 \log_2 x} \cdot 2^{\ell} \log_3 2 + k} = (2^{2^{\log_2 \log_2 x}})^{2^{\ell} \log_3 2 + k} = x^{2^{\ell} \log_3 2 + k} \end{align*}$$
$$\text{so } 2^{\ell \log_3 2+k} = 2 \iff \ell \log_3 2 + k = 1.$$
$$\begin{align*} g(f(x)) &= g(2^{2^{\log_3 2 \log_2 \log_2 x+k}}) = 2^{2^{\log_2 3 \log_2 \log_2 2^{2^{\log_3 2 \log_2 \log_2 x+k}}}+ \ell} = 2^{2^{\log_2 3 \cdot (\log_3 2 \log_2 \log_2 x+k) + \ell}} \\ &= 2^{2^{\log_2 \log_2 x+k} \log_2 3+\ell} = 2^{2^{\log_2 \log_2 x} \cdot 2^k \log_2 3+\ell} = (2^{2^{\log_2 \log_2 x}})^{2^k \log_2 3+\ell} = x^{2^k \log_2 3+\ell} \end{align*}$$
$$\text{so } 2^{k \log_2 3 + \ell} = 3 \iff k \log_2 3 + \ell = \log_2 3.$$
It is clear that both $f$ and $g$ are well defined in $S$ and that it's possible to
choose $k$ and $\ell$ (for example, $k=1$ and $\ell=0$), so such functions do exist.