First, you must compute the derivative of y(x), which is $y'(x)=3x^2-12x+15$, and so $y'(3)=6$, a value that will be the slope of the tangent line to $y(x)$ at the point $P=(3,9)$. Now we use the following expression for a line, where $(x_0,y_0)$ is a point of the line (in this case $P=(3,9)$) and $m$ is the slope at that point (in this case $m=6$):
we conclude that the equation of the tangent line we are looking for is $y-9=6(x-3)$, that is, $y=6x-9$. We must see that, if $N$ is the foot of the perpendicular from $P$ to the $x$-axis, then the previous tangent line passes through the mid-point of $ON$, so let's calculate the explicit expression of these elements. The foot will be obviously the point $(3,0)$, and since the point coordinates of the origin are $(0,0)$, the midpoint of $ON$ will be $Q=(3/2,0)$.
And does $Q$ belong to the tangent line? Yes, because $6\cdot(3/2)-9=0$, thus completing the proof.