Let $y = f(x)$ be the real valued function defined on the interval $(0, \infty)$, satisfying $y(1) = 0$ and the differential equation
$x\dfrac{dy}{dx} = y - x^3$.
Then which of the following statements is (are) TRUE ?

The given differential equation is $x\dfrac{dy}{dx} = y - x^3$.

Dividing by $x$, we get:
$\dfrac{dy}{dx} - \dfrac{1}{x}y = -x^2$

This is a linear differential equation of the form $\dfrac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = -\dfrac{1}{x}$ and $Q(x) = -x^2$.

Integrating Factor (IF) = $e^{\int -\dfrac{1}{x} dx} = e^{-\ln x} = \dfrac{1}{x}$.

Multiplying the differential equation by the IF:
$\dfrac{d}{dx} \left( y \cdot \dfrac{1}{x} \right) = -x$

Integrating both sides with respect to $x$:
$\dfrac{y}{x} = -\dfrac{x^2}{2} + C$
$y = -\dfrac{x^3}{2} + Cx$

Given $y(1) = 0$:
$0 = -\dfrac{1}{2} + C \Rightarrow C = \dfrac{1}{2}$

Thus, $f(x) = \dfrac{x}{2} - \dfrac{x^3}{2}$.

To find local extrema, we find $f'(x)$:
$f'(x) = \dfrac{1}{2} - \dfrac{3x^2}{2}$

Setting $f'(x) = 0$ gives $3x^2 = 1 \Rightarrow x = \dfrac{1}{\sqrt{3}}$ (since $x > 0$).

Now, $f''(x) = -3x$. At $x = \dfrac{1}{\sqrt{3}}$, $f''\left(\dfrac{1}{\sqrt{3}}\right) = -\sqrt{3} < 0$.
So, $f(x)$ has a local maximum at $x = \dfrac{1}{\sqrt{3}}$.

For $x \in (1, 2)$, $x^2 > 1$, so $f'(x) = \dfrac{1 - 3x^2}{2} < 0$. Thus, $f(x)$ is decreasing in $(1, 2)$.

For the intersection of $f(x)$ and $g(x)$:
$\dfrac{x}{2} - \dfrac{x^3}{2} = 4x^3 - 5x^2 + \dfrac{3}{2}x$
$x - x^3 = 8x^3 - 10x^2 + 3x$
$9x^3 - 10x^2 + 2x = 0$
$x(9x^2 - 10x + 2) = 0$

Since $x \in (0, \infty)$, $x \neq 0$. We solve $9x^2 - 10x + 2 = 0$.
Discriminant $\Delta = (-10)^2 - 4(9)(2) = 100 - 72 = 28 > 0$.
Sum of roots $= \dfrac{10}{9} > 0$ and Product of roots $= \dfrac{2}{9} > 0$.
Both roots are real and positive. Therefore, there are exactly $2$ elements in the set.

Answer: The function $f$ has a local maximum at $x = \dfrac{1}{\sqrt{3}}$; If $g(x) = 4x^3 - 5x^2 + \dfrac{3}{2}x$ for $x > 0$, then the number of elements in the set $\{x \in (0, \infty) : f(x) = g(x)\}$ is $2$