How many satisfied values?

Let's change variables for clarity, and solve $|f(t^3-3t)| = \frac23.$

First let's look at the solutions to $|f(x)| = \frac23.$ Drawing horizontal lines at $y=\pm 2/3$ in the above graph, there are six $x$ -coordinates of intersection. One of them is in the interval $(-\infty, -2),$ two of them are in the interval $(-2,2),$ and three of them are in the interval $(2,\infty).$ So $t^3-3t$ is one of these six values.

Now drawing the graph of $x=t^3-3t$ and doing the same horizontal line test shows that there are three values of $t$ for any $x$ in $(-2,2),$ and one value of $t$ for any $x$ in $(-\infty,-2)$ and any $x$ in $(2,\infty).$ So the total number of solutions is $3 \cdot 2 + 1 \cdot 4 = \fbox{10}.$

From the graph, we know three things:
1) The cubic has a root at $x = 2$
2) The cubic has a root at $x = -2$
3) The function has a value of $-1$ when the derivative of the function is zero

The following form results:

$f(x) = (x-2)(x+2)(x-a) \\ f(x) = x^3 - ax^2 - 4x + 4a \\ f'(x) = 3x^2 - 2ax - 4$

Find an $(x,a)$ pair which solves the following equations:

$-1 = x^3 - ax^2 - 4x + 4a \\ 0 = 3x^2 - 2ax - 4$

Solving using multi-variate Newton Raphson gives:

$(x,a) \approx (2.4969, 2.9444)$

The plot of $f(x)$ is below:

The plot of $f(x^3 - 3x)$ is below. The magnitude is equal to $\frac{2}{3}$ ten times