Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a function defined by
$f(x) = \begin{cases} x\left(\dfrac{e^{1/x} - e^{-1/x}}{e^{1/x} + e^{-1/x}}\right) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0. \end{cases}$
Then

The given function can be rewritten as $f(x) = x \tanh\left(\dfrac{1}{x}\right)$ for $x \neq 0$, and $f(0) = 0$.

Let us check the continuity of $f$ at $x = 0$:
$\lim_{x \to 0} f(x) = \lim_{x \to 0} x \tanh\left(\dfrac{1}{x}\right)$
Since $\left|\tanh\left(\dfrac{1}{x}\right)\right| < 1$ for all $x \neq 0$, we have $|f(x)| \leq |x|$.
As $x \to 0$, $|f(x)| \to 0$, which means $\lim_{x \to 0} f(x) = 0 = f(0)$. Thus, $f$ is continuous at $x = 0$.

Let us check the differentiability of $f$ at $x = 0$:
Right-hand derivative: $f'(0^+) = \lim_{h \to 0^+} \dfrac{h \tanh(1/h) - 0}{h} = \lim_{h \to 0^+} \tanh\left(\dfrac{1}{h}\right) = 1$
Left-hand derivative: $f'(0^-) = \lim_{h \to 0^-} \dfrac{h \tanh(1/h) - 0}{h} = \lim_{h \to 0^-} \tanh\left(\dfrac{1}{h}\right) = -1$
Since $f'(0^+) \neq f'(0^-)$, $f$ is not differentiable at $x = 0$.

Let us check the monotonicity of $f$ in $(-1, 1)$:
$f(-x) = (-x) \tanh\left(-\dfrac{1}{x}\right) = x \tanh\left(\dfrac{1}{x}\right) = f(x)$
Since $f$ is an even function and non-constant, it cannot be monotonically increasing over the interval $(-1, 1)$.

Let us check if $f$ is bounded:
For $x \neq 0$, let $t = \dfrac{1}{x}$. Then $f(x) = \dfrac{\tanh t}{t}$.
We know that for all $t \in \mathbb{R}$, $|\tanh t| \leq |t|$, which implies $\left|\dfrac{\tanh t}{t}\right| \leq 1$ for $t \neq 0$.
Therefore, $|f(x)| \leq 1$ for all $x \neq 0$. Since $f(0) = 0$, we have $|f(x)| \leq 1$ for all $x \in \mathbb{R}$.
Thus, there exists a constant $C = 1$ such that $|f(x)| \leq C$ for all $x \in \mathbb{R}$.

Answer: there exists a constant $C$ such that $|f(x)| \leq C$ for all $x \in \mathbb{R}$.