Find the point on the curve $y=x^3-3x$ at which tangent is parallel to $X$-axis.

 Let the point at which tangent is parallel to $X$-axis be $P\left(x_1,y_1\right)$.

Then, it must lie on curve.

Therefore, we have $y_1=x_1^3-3x_1$

Differentiating $y=x^3-3x$ w.r.t. $x$, we get

$\frac{dy}{dx}=3x^2-3\Rightarrow {\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}=3{x_1}^2-3$

Since, the tangent is parallel to $X$-axis.

${\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}=0\Rightarrow 3{x_1}^2-3=0$

$x_1=\pm 1$

When $x_1=1$, then $y_1=1-3=-2$

When $x_1=-1$, then $y_1=-13=2$

$\therefore$ Points at which tangent is parallel to $X$-axis are $(1,-2)$ and $(-1,2)$.