(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.
The only assumption on $f$ which is identifiable in your question seems to be that it has a directional derivative at some point $w_0 \in l$ in the direction of a unit tangent $u$ to $l$ at $w_0$. Let us call this the minimal assumption. We shall see that it is too weak to get the desired result.
For a general function $f$ the level set $l = f^{-1}(k)$ may be any subset of $\mathbb R^2$. Even if $f$ is assumed to be continuous, $l$ may be any closed subset of $\mathbb R^2$, thus it still may be really weird. We must not think it is a $1$-dimensional $C^1$-submanifold of $\mathbb R^2$. If it is, this would be a strong additional assumption on $f$. A convenient condition assuring this is that $f$ is smooth and $k$ is a regular value of $f$. However, this would be far too restrictive.
Let us first clarify what it means to consider a tangent vector to $l$. In the general case the only reasonable interpretation seems to be the following:
Let $w_0 = (x_0,y_0) \in l$ and $\phi : (-a,a) \to \mathbb R^2$ be a $C^1$-curve such that $\phi(0) = w_0$ and $\phi((-a,a)) \subset l$. Then $\phi'(0)$ is a tangent vector to $l$ at $w_0$. If $\phi'(0) \ne 0$ we may re-parameterize $\phi$ to get a unit tangent vector. There may be many unit tangent vectors to $l$ at $w_0$. But if $l$ is a $1$-dimensional $C^1$-submanifold of $\mathbb R^2$, there is only one up to sign. It may also happen that zero is the only tangent vector, for example if $w_0$ is an isolated point of $l$.
The following two examples show that the minimal assumption is too weak.
Example 1: $$f(x,y) = \begin{cases}
0 & y = x^2 \\
x & y = 0 \\
1 & \text{else}
\end{cases}$$
This is a non-continuous function whose level set $l = f^{-1}(0)$ is the parabola $y = x^2$ (in particular it is a $1$-dimensional $C^1$-submanifold of $\mathbb R^2$). A unit tangent vector at $(0,0)$ to $P$ is $(1,0)$, and $f$ has a directional derivative at $(0,0)$ in direction $(1,0)$. We have $f(x,0) = x$, thus this directional derivative has the value $1 \ne 0$.
Example 2: Let $\Delta$ be the set of all $(x,y)$ such that $\lvert y \rvert = x^2$. This "double" parobola is not a $1$-dimensional $C^1$-submanifold of $\mathbb R^2$.
$$f(x,y) = \begin{cases}
x - \frac{\lvert y \rvert}{x} & x \ne 0, \lvert y \rvert \le x^2 \\
0 & x = y = 0 \\
d(\Delta,(x,y)) & \text{else}
\end{cases}$$
Here $d(\Delta,(x,y))$ denotes the Euclidean distance of $(x,y)$ to $\Delta$. Clearly $f$ is a continuous function whose level set $f^{-1}(0)$ is $\Delta$. It has two unit tangent vectors at $(0,0)$. Taking $(1,0)$, we see that $f$ has a directional derivative at $(0,0)$ in this direction. We have $f(x,0) = x$, thus this directional derivative has the value $1 \ne 0$.
I think these examples show that some differentiability assumption is necessary to obtain the desired result.
So let us assume that $f$ is differentiable at $w_0$ with derivative $df(w_0)$ (which is a linear map).
Then the directional derivative of $f$ at $w_0$ in any direction $\omega \in \mathbb R^2$ exists and has the value $df(w_0)(\omega)$. This is of course much stronger than assumimg that some directional derivative of $f$ exists.
Now let $\phi : (-a,a) \to \mathbb R^2$ be a $C^1$-curve such that $\phi(0) = w_0$ and $\phi((-a,a)) \subset l$. Its tangent vector at $w_0$ is $\omega = \phi'(0)$. We claim that $df(w_0)(\omega) = 0$ which is the desired result. It is trivial for $\omega = 0$. If $\omega \ne 0$, we know that $\lVert \frac{\phi(t) - \phi(0)}{t} \rVert > 0$ for $\lvert t \rvert < \epsilon$. Thus $\phi(t) \ne \phi(0)$ for $\lvert t \rvert < \epsilon$. We know that
$$\lim\limits_{w \to w_0} \frac {f(w) - f(w_0) - df(w_0)(w-w_0)} {\lVert w - w_0 \rVert} = 0 .$$
This implies
$$\lim\limits_{t \to 0} df(w_0)\left(\frac{\phi(t) -\phi(0)}{\lVert \phi(t) -\phi(0) \rVert} \right) = \lim\limits_{t \to 0} \frac {f(\phi(t)) - f(\phi(0)) - df(w_0)(\phi(t) -\phi(0))} {\lVert \phi(t) -\phi(0) \rVert} = 0 .$$
We know that $\lim\limits_{t \to 0} \frac{\phi(t) -\phi(0)}{t} = \omega$, thus $\lim\limits_{t \to 0} \lVert \frac{\phi(t) -\phi(0) }{t} \rVert = \lVert \omega \rVert$ and $\lim\limits_{t \to 0} \frac{\phi(t) -\phi(0)}{\lVert \phi(t) -\phi(0) \rVert} = \frac{\omega}{\lVert \omega \rVert}$. Hence
$$0 = \lim\limits_{t \to 0} df(w_0)\left(\frac{\phi(t) -\phi(0)}{\lVert \phi(t) -\phi(0) \rVert} \right) = df(w_0)\left(\lim\limits_{t \to 0}\frac{\phi(t) -\phi(0)}{\lVert \phi(t) -\phi(0) \rVert} \right) = df(w_0)\left(\frac{\omega}{\lVert \omega \rVert}\right) \\ = \frac{1}{\lVert \omega \rVert}df(w_0)(\omega) .$$
Best Answer
$z=\pm 5$ is not a curve. Of course, $2x^2+2y^2-z^2=25$ and $x^2+y^2=z^2$ imply $z=\pm 5$, but don't forgot about $x$ and $y$. I'll show case $z=5$, $z=-5$ is similar.
If $z=5$, then $x^2+y^2=25$, so you can parametrize curve this way:
$$(x,y,z)=(5\sin t, 5\cos t, 5)$$
Now put $x=5\sin t, y=5\cos t, z=5$ into $f(x,y,z) = x^2+y^2-z^2$. You get function of one variable:
$$g(t)=f(x,y,z)=25\sin^2 t+ 25\cos^2 t -25=0$$
Finally calculate derivative of this function. It's directional derivative which you find.
$$g'(t)=0$$