Finding an equation of a plane perpindicular to xy plane that intersects with a surface and has a directional derivate of zero at this point. I'm a bit new to 3D space and haven't had much practice with it. One question I'm working on says:

A plane perpendicular to the x-y plane contains the point (3, 2, 2) on
  the paraboloid $36z=4x^2+9y^2$. The cross-section of the paraboloid
  created by this plane has slope 0 at this point. Find an equation of
  the plane.

What I did was I solved for z explicitly, giving me:
$z=\frac{4}{36}x^2 + \frac{9}{36}y^2$ 
Then I found the partials with respect to $x$ and $y$ giving me: 
$F_x = \frac{8}{36}x$ 
$F_y = \frac{18}{36}y$ 
Evaluating the partials at the point $(3,2,2)$ gives me: 
$F_x(3) = \frac{24}{34}$ 
$F_y(2) = 1$ 
Since it's given that the directional derivative at this point is zero, and that the plane is perpendicular to the x-y plane, I get the equation of the plane to be: 
$0=\frac{24}{34}(x-3) + (y-2)$ 
But the answer in the textbook says: 
$0=2(x-2)+3(y-2)$ 
Any help would be greatly appreciated. 
Thanks
 A: Our plane intersects the $xy$ plane in a straight line that goes throught the point $(3,2)$ and whose slope is unknown. Let $m$ be the slope. So, the equation of the intersection line is
$$y=mx-3m+2.$$
If we subtitute this $y(x)$ into the equation of the paraboloid we'll get the equation of the intersection of the plane and the paraboloid (projected to the $xz$ plane):
$$z=\frac19x^2+\frac14(mx-3m+2)^2.$$
The derivative of $z(x)$ with respect to $x$ equals $0$ at $x=3$ and $z=2$ because the tangent line in the projection looks the same as it does in the space. So
$$\frac{\operatorname dz}{\operatorname dx}_{\text{ at }x=3}=\left[\frac29x+\frac m2(mx-3m+2)\right]_{\text{ at }x=3}=0.$$
That is,
$$m=-\frac23.$$
Substituting this $m$ back to the straight line's equation we get
$$y=-\frac23x+4.$$
Multiplying both sides of the equation above by $3$ we get
$$3y+2x-12=0$$
or
$$3y-6+2x-6=0$$
or
$$3(y-2)+2(x-3)=0$$
which is the equation of the plane (at the same time) because the plain is perpendicular to the $xy$ plane.
