The number of points, at which the function f(x) = \ 6x, 2 + 3x^2\...

Let $f(x) = g(x) + h(x)$, where
$g(x) = \max\{6x,\ 2 + 3x^2\}$
$h(x) = |x - 1|\left|\cos\left|x^2 - \dfrac{1}{4}\right|\right|$

We find the points of non-differentiability of each part in $x \in (-\pi, \pi)$.

Non-differentiability of $g(x)$:
$g(x) = \max\{6x,\ 2 + 3x^2\}$ is non-differentiable where the two curves intersect with different slopes.

$6x = 2 + 3x^2 \Rightarrow 3x^2 - 6x + 2 = 0$
$x = \dfrac{6 \pm \sqrt{36 - 24}}{6} = \dfrac{6 \pm 2\sqrt{3}}{6} = 1 \pm \dfrac{1}{\sqrt{3}}$

Both values lie in $(-\pi, \pi)$. At these points, the derivatives of the two inner functions are $6$ and $6x$; since $x \neq 1$, the slopes differ.
So $g(x)$ contributes $2$ points.

Non-differentiability of $h(x)$:
Since $\cos|y| = \cos y$:
$h(x) = |x - 1|\left|\cos\left(x^2 - \dfrac{1}{4}\right)\right|$

An expression of the form $|u(x)|$ is non-differentiable at simple zeros of $u(x)$.

Case A: $x - 1 = 0 \Rightarrow x = 1$
At $x = 1$: $\cos\left(1 - \dfrac{1}{4}\right) = \cos\dfrac{3}{4} \neq 0$, so $x = 1$ is a point of non-differentiability.
Contribution: $1$ point.

Case B: $\cos\left(x^2 - \dfrac{1}{4}\right) = 0$
$x^2 - \dfrac{1}{4} = (2n + 1)\dfrac{\pi}{2} \Rightarrow x^2 = \dfrac{1}{4} + (2n + 1)\dfrac{\pi}{2}$

Domain restriction: $x \in (-\pi, \pi) \Rightarrow x^2 \in [0, \pi^2)$, and $\pi^2 \approx 9.87$.

$n = -1$: $x^2 = 0.25 - \dfrac{\pi}{2} < 0$ (rejected)
$n = 0$: $x^2 \approx 1.82 < 9.87$ (gives $2$ points)
$n = 1$: $x^2 \approx 4.96 < 9.87$ (gives $2$ points)
$n = 2$: $x^2 \approx 8.10 < 9.87$ (gives $2$ points)
$n = 3$: $x^2 \approx 11.25 > 9.87$ (rejected)

Contribution: $2 + 2 + 2 = 6$ points.

Checking for overlap:
The points $x = 1 \pm \dfrac{1}{\sqrt{3}}$ are not equal to $1$, and for these, $x^2 - \dfrac{1}{4} = \dfrac{13}{12} \pm \dfrac{2}{\sqrt{3}}$, which are not odd multiples of $\dfrac{\pi}{2}$. So the non-differentiable points of $g(x)$ and $h(x)$ are disjoint.

Total number of points of non-differentiability:
$2 + 1 + 6 = 9$

Hence, the answer is $9$.