Let $f(x) = \lim_{y \to 0} \dfrac{(1 - \cos(xy)) \tan(xy)}{y^3}$. Then the number of solutions of the equation $f(x) = \sin x$, $x \in \mathbf{R}$ is :

The given function is $f(x) = \lim_{y \to 0} \dfrac{(1 - \cos(xy)) \tan(xy)}{y^3}$.

Multiplying and dividing by $x^3$, we get:

$f(x) = \lim_{y \to 0} \dfrac{1 - \cos(xy)}{(xy)^2} \cdot \dfrac{\tan(xy)}{xy} \cdot x^3$

Using the standard limits $\lim_{t \to 0} \dfrac{1 - \cos t}{t^2} = \dfrac{1}{2}$ and $\lim_{t \to 0} \dfrac{\tan t}{t} = 1$, we obtain:

$f(x) = \dfrac{1}{2} \cdot 1 \cdot x^3 = \dfrac{x^3}{2}$

We need to find the number of solutions for the equation $f(x) = \sin x$, which is $\dfrac{x^3}{2} = \sin x$.

Let $g(x) = \dfrac{x^3}{2} - \sin x$.

Differentiating with respect to $x$:

$g'(x) = \dfrac{3x^2}{2} - \cos x$

$g''(x) = 3x + \sin x$

For $x > 0$, $g''(x) > 0$, which implies that $g'(x)$ is strictly increasing on $(0, \infty)$.

We have $g'(0) = -1 < 0$ and $g'(\pi/2) = \dfrac{3\pi^2}{8} > 0$. Thus, $g'(x) = 0$ has exactly one root in $(0, \infty)$, say at $x = x_0$.

This means $g(x)$ decreases on $(0, x_0)$ and increases on $(x_0, \infty)$.

Since $g(0) = 0$ and $g(x)$ initially decreases, $g(x_0) < 0$. As $x \to \infty$, $g(x) \to \infty$. Therefore, $g(x) = 0$ has exactly one positive root.

Since $g(x)$ is an odd function ($g(-x) = -g(x)$), it must also have exactly one negative root.

Including the trivial root $x = 0$, the equation $g(x) = 0$ has exactly $3$ solutions.

Answer: $3$