Evaluate the directional derivative of $f$ for the points and directions specified Evaluate the directional derivative of $f$ for the points and directions specified:
(a) $f(x,y,z)=3x-5y+2z$ at $(2,2,1)$ in the direction of the outward normal to the sphere $x^2+y^2+z^2=9.$
(b) $f(x,y,z)=x^2-y^2$ at a general point of the surface $x^2+y^2+z^2=4$ in the direction of the outward normal at that point.
(c) $f(x,y,z)=x^2+y^2-z^2$ at $(3,4,5)$ along the curve of intersection of the two surfaces $2x^2+2y^2-z^2=25$ and $x^2+y^2=z^2.$
The answer for the problems are (a) -2/3 (b) $x^2-y^2$ (c) $0$.
For (a) and (b) I thought the direction of the outward normal was the gradient of the level surface, which would be $(2x,2y,2z)$ in both cases. So to get the directional derivative I would simply have to compute the dot product of grad$f$ and $(2x,2y,2z)$ at the corresponding points, however, the result does not match the answers. 
For (c), I'm stuck on finding the curve of intersection of the two surfaces.
I would greatly appreciate it if anyone could help me out.

(a)
The outward normal of the sphere given is $\left(\frac{x}{3},\frac{y}{3}, \frac{1}{3}\sqrt{9-x^2-y^2}\right)$. The same at the given point is $\overline n=\left(\frac{2}{3},\frac{2}{3}, \frac{1}{3}\right).$ The vector of the partial derivatives is $\nabla f =(3,-5,2).$ The directional derivative is then
$$\nabla f \cdot \overline n=\left(\frac{2}{3},\frac{2}{3}, \frac{1}{3}\right)\cdot (3,-5,2)=2-\frac{10}{3}+\frac{2}{3}=-\frac{2}{3}.$$
(b)
The outward normal of the sphere given is $\left(\frac{x}{2},\frac{y}{2}, \frac{1}{2}\sqrt{4-x^2-y^2}\right)$.The vector of the partial derivatives is $\nabla f =(2x,-2y,0)$. So,
$$\nabla f \cdot \overline n=(2x,-2y,0)\cdot\left(\frac{x}{2},\frac{y}{2}, \frac{1}{2}\sqrt{4-x^2-y^2}\right)=x^2-y^2.$$
(c)
We have two surfaces this time: $2x^2+2y^2-z^2=25$ and $x^2+y^2=z^2.$ These intersect above the circle $x^2+y^2=25$ at $z=\pm 5$. (The intersection line was given by setting $2x^2+2y^2-25=x^2+y^2$. Also, where $x^2+y^2=25$, there $z=\pm 5$.) Now, the vectors pointing to the intersection lines are of the form $(x,\sqrt{25-x^2},\pm 5)$; at $(3,4,5)$ it gives $(3,4,5)$. The vector of the partial derivatives: $(2x,2y,-2z)$, which at the given point is $(6,8,-10)$. So,
$$\nabla f \cdot \overline n= (3,4,5)\cdot(6,8,-10)=0.$$
(c')
I've seen that the other people answering considered the tangent vector to the intersection line. That approach gave the same result at the given point. If "along" means "tangent to" then my answer was only accidentally correct. 
