The number of points on the curve $y=x^3-4x$ at which the tangent is parallel to the $x$-axis are:

Let the required point where the tangent is parallel to $x$-axis as $(x_1,y_1)$.

Substituting in the curve,

$y_1={x_1}^3-4x_1$

Differentiating the above equation on both sides with respect to $x$,

$\frac{d{y}_1}{d{x}_1}=3{x_1}^2-4$

The value $\frac{d{y}_1}{d{x}_1}$ is the slope of the tangent to the curve at $(x_1,y_1)$.

We know that the slope of the line parallel to $x$-axis is $0$ that is,

$$\begin{gathered} \Rightarrow 3{x_1}^2-4=0 \\ \Rightarrow {x_1}^2=\frac{4}{3} \\ \Rightarrow x_1=\pm \sqrt{\frac{4}{3}} \\ \Rightarrow x_1=\pm \frac{2}{\sqrt{3}} \end{gathered}$$

There are two possible values for $x_1$

When $x_1=+\frac{2}{\sqrt{3}}$ the value of ordinate is,

$$\begin{gathered} y_1={\left(\frac{2}{\sqrt{3}}\right)}^3-4\left(\frac{2}{\sqrt{3}}\right) \\ {y}_1=\frac{8}{3\sqrt{3}}-\frac{8}{3\sqrt{3}} \\ {y}_1=0 \end{gathered}$$

Similarly, when $x_1=-\frac{2}{\sqrt{3}}$,

$$\begin{gathered} y_1={\left(-\frac{2}{\sqrt{3}}\right)}^3-4\left(-\frac{2}{\sqrt{3}}\right) \\ {y}_1=-\frac{8}{3\sqrt{3}}+\frac{8}{3\sqrt{3}} \\ {y}_1=0 \end{gathered}$$

So, the points are $\left(\frac{2}{\sqrt{3}},0\right)$ and $\left(-\frac{2}{\sqrt{3}},0\right)$.

So, there are $2$ points on the curve where the tangent is parallel to $x$-axis.