Find the points on the curve $y = {x^3}$ at which the slope of the tangent is equal to the $y$-coordinate of the point.

We have,
$y = {x^3}$ …(i)

Differentiating (i), w.r.t. $x$ , we get
$\cfrac{{dy}}{{dx}} = 3{x^2}$
Since, it is given that slope is equal to the $y -$coordinate of the point

$\therefore \cfrac{{dy}}{{dx}} = y \Rightarrow 3{x^2} = y$ (using (ii))
$\Rightarrow 3{x^2} = {x^3} \Rightarrow {x^2}\left( {3 - x} \right) = 0 \Rightarrow x = 0$ or $x = 3$ (using (i))

When $x = 0$, then from (i)$y = 0$
When $x = 3$, then from (i), $y = {3^3} = 27$

Therefore the required points are $(0,0)$ and $(3,\;27)$.