Relation $R$: $R = \{(x, y) | x, y$ are real numbers and $x = wy$ for some rational number $w\}$.

Reflexivity: For all $x$ in $R$, $x = 1 \cdot x$. Since $1$ is a rational number, every element is related to itself.
Symmetry: For all $x, y$ in $R$, if $x = wy$ for some rational $w$, then $y = \frac{1}{w}x$. However, if $w = 0$, then $\frac{1}{w}$ is undefined, and therefore, $R$ doesn't satisfy symmetry.
Transitivity: If $x = wy$ and $y = vz$ for some rational numbers $w$ and $v$, then $x = (wv)z$. Since the product of rational numbers is rational, if $x$ is related to $y$ and $y$ is related to $z$, then $x$ is related to $z$.

Therefore, $R$ is not an equivalence relation on $R$ since it does not satisfy the symmetry property.

Relation $S$: $S = \{(\frac{m}{n}, \frac{p}{q}) | m, n, p$ and $q$ are integers such that $n, q \neq 0$ and $qm = pn\}$.

Reflexivity: For all $\frac{m}{n}$, $\frac{m}{n} = \frac{m}{n}$. Since $n \neq 0$ and $m = m$, every element is related to itself.
Symmetry: For all $\frac{m}{n}, \frac{p}{q}$, if $qm = pn$, then $np = mq$. So if $\frac{m}{n}$ is related to $\frac{p}{q}$, then $\frac{p}{q}$ is related to $\frac{m}{n}$.
Transitivity: If $\frac{m}{n}R\frac{p}{q}$ and $\frac{p}{q}R\frac{r}{s}$ $\Rightarrow mq = np$ and $ps = rq$ $\Rightarrow mq \cdot ps = np \cdot rq$ $\Rightarrow ms = nr$ $\Rightarrow \frac{m}{n} = \frac{r}{s}$ $\Rightarrow \frac{m}{n}R\frac{r}{s}$.

So if $\frac{m}{n}$ is related to $\frac{p}{q}$ and $\frac{p}{q}$ is related to $\frac{r}{s}$, then $\frac{m}{n}$ is related to $\frac{r}{s}$. The relation $S$ is transitive.

Therefore, $S$ is an equivalence relation on the set of all fractions where the denominator is not zero.

In conclusion, $S$ is an equivalence relation but $R$ is not an equivalence relation.