Let $(x,y)$ be a point on the elliptic curve $E$ given by $y^2=x^3+Ax+B$. Show that if $y=0$ then $3x^2+A\ne0$.
I have a graphical intuition for this. Since $$\frac{dy}{dx}=\frac{3x^2+A}{2y}$$ and tangent to the elliptic curve at the point $(x,0)$ is parallel to $y-axis$. So it makes that the numerator of the slope at that point is non-zero and the denominator is zero. But I'm not able to mathematically prove this. Any help is greatly appreciated.
If you set $F(x,y)=y^2-f(x)$, then your elliptic curve is the points that satisfy the equation $F(x,y)=0$.
Now $F$ is sooth iff it is non-singular, that is, if there exists no point $(x_0,y_0)$ that satisfies $F(x_0,y_0)=0$ at which both first order partial derivatives vanish.
The partial derivatives of $F$ are:
$$F_x(x,y)=-f'(x)$$ and $$F_y(x,y)=2y$$
If we have the point $(x,0)$, then $y=0$ implies that $F_y(x,0)=0$ and hence by our remark above, in order for our curve to be smooth, we must have that: $$F_x(x,0)\neq 0\Leftrightarrow -f'(x)\neq 0 \Leftrightarrow f'(x)\neq 0$$ which is precisely what you want to show, if you substitute $f(x)=x^3+Ax+B$