Step 1:

Let $P(x_1,y_1)$ be the required point.

The given curve is $y=x^3$----(1)

Differentiating w.r.t $x$ we get,

$\large\frac{dy}{dx}$$=3x^2$

The slope of the tangent at $P(x_1,y_1)$ is

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

Step 2:

It is given that the slope of the tangent is equal to the $y$ coordinate of the point.

Hence $\large\frac{dy}{dx}$$=y_1$------(3)

Equating equ(2) and equ(3) we get,

$3x_1^2=y_1$-----(4)

Also $(x_1,y_1)$ lies on equ(1)

Hence $y_1=x_1^3$----(5)

Step 3:

Now equating equ(4) and equ(5)

$3x_1^2=x_1^3$

$\Rightarrow 3x_1^2-x_1^3=0$

$x_1^2(3-x_1)=0$

Therefore $x_1=0$ and $x_1=3$

Step 4:

When $x_1=0,y_1=(0)^3=0$

When $x_1=3,y_1=(3)^3=27$

Hence the required points are $(0,0)$ and $(3,27)$