Apostol Calculus Vol 2 Exercise 8.17 Q No. 3 Evaluate the directional derivative of $f$ for the points and directions specified
$f(x,y,z)=x^2+y^2-z^2$ 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 is supposed to be $0.$ I am not getting the right answer maybe because I don't know how to get the curve of intersection or I don't really understand the directional derivative enough. This is what I did though:
\begin{align}
2x^2+2y^2-z^2&=25\\
\implies 2(x^2+y^2)-z^2&=25\\
\implies z^2&=25
\end{align}
Thus the curve of intersection is $x^2+y^2=25.$ which is a circle centred at orgin with radius $5.$ Hence if $(x,y)$ is any point of the circle, then the unit direction towards that point is $\frac{1}{5}(x,y).$ Also $\nabla f=(2x,2y,-2z)$ and hence we have $\nabla f(3,4,5)=(6,8,-10)$. 
Thus the directional derivative at the point $(3,4,5)$ is $\nabla f(3,4,5)\cdot \big(\frac{3}{5},\frac{4}{5}\big)=\frac{58}{5}.$(a wrong answer)

I think your unit direction vector should be  : $\left(
\begin{array}{c}
-\frac{4}{5}\\
\frac{3}{5}\\
0\\
\end{array}
\right)$
 (maybe I'll include picture later).
And that the answer should be : $0$ accordingly : $\left(0=\left(
\begin{array}{c}
-\frac{4}{5}\\
\frac{3}{5}\\
0\\
\end{array}
\right) \cdot \left(
\begin{array}{c}
6\\
8\\
-10\\
\end{array}
\right)\right)$.
Here a picture of the tangent line in $(3,4)$   : 
