It means, along the tangent vector to the curve.
There is a very simple way to do this: substitute $x,y,z$ with the coordinates of the curve, getting:
$$
F(t) = f(x(t),y(t),z(t)) = f(e^{-t}, 1+2\sin t, t-\cos t) = ...
$$
Now simply take the derivative with respect to $t$!
EDIT Otherwise, take the scalar product between the gradient of $f$, and the tangent vector to the curve at $t=0$.
The tangent vector to the curve is the vector:
$$
v= \bigg(\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}\bigg),
$$
so in our case, $(-e^{-t},etc.)$.
If your book (or professor) want you to take a unit tangent vector, divide this by the norm (but I don't think you have to do it, along the curve...ask your teacher).
The gradient is:
$$
\nabla f = \bigg(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\bigg),
$$
so in our case, $(2xyz^3,etc.)$.
Now take the scalar product of these two vectors, at $t=0$. Which means at the coordinates $(x(0),y(0),z(0))= (1,etc.)$
Best Answer
You can do this without finding an intersection. Just find normal vectors to both surfaces (i.e. compute gradients) and then compute theirs cross product to obtain curve tangent vector.
For the first surface the gradient is $$ (4x, 4y, -2z). $$ The second one gives us $$ (2x, 2x, -2z). $$
So, we can take $(2x, 2y, -z)$ and $(x, y, -z)$ as normal vectors and compute their cross product to obtain tangent vector $$ \boldsymbol{v} = \begin{vmatrix} \boldsymbol{i} & \boldsymbol{j} &\boldsymbol{k} \\ 2x & 2y & -z \\ x & y & -z \end{vmatrix} = -yz \boldsymbol{i} +xz\boldsymbol{j}. $$
Let's substitute $(3,4,5)$ $$ \boldsymbol{v} = -4\cdot 5\boldsymbol{i} + 3\cdot{5}\boldsymbol{j}. $$ It's easy to see that normalizing this vector we get $$ \boldsymbol{v}_1 = \frac{\boldsymbol{v}}{|\boldsymbol{v}|} = -\frac{4}{5} \boldsymbol{i} + \frac{3}{5}\boldsymbol{j}. $$
Now you need to find the directional derivative along $\boldsymbol{v}_1$ or $-\boldsymbol{v}_1$ (both will be tangent vectors indicating two different directions on the curve).
The other way for solving this problem (as suggested in the comments) was to eliminate $z$ from the equations and get the curve (WLOG we take positive $z$ considering the case $z = -5\sqrt{2}$ by analogy) $$ x^2 + y^2 = 50, \; z = 5\sqrt{2}. $$ So, we see that the curve is a circle $x^2 + y^2 = 50$ located at the "height" $z=5\sqrt{2}$ (this actually explains why $k$ coordinate is equal to zero in tangent vector). The tangent vector to the circle $$ x^2 + y^2 = 50 $$ is equal to $(-y, x)$ and after substitution we get again $$ \boldsymbol{v} = (-4, 5) \implies \boldsymbol{v}_1 = \frac{\boldsymbol{v}}{|\boldsymbol{v}|} = -\frac{4}{5} \boldsymbol{i} + \frac{3}{5}\boldsymbol{j}. $$
All you need now is to compute the directional derivative (i.e. to find dot product of gradient of the function and obtained unit vector).