Finding directional derivative in direction of tangent of curve just something small I couldnt get. $C$ is my curve that described by intersection of two planes: $$2x^2-y^2\:=1 ,\:2y-z=0 $$
The point $(1,1,2)$ is on the curve. $n$ is the vector whos direction is the tangent to $C$ in the point (in a way that its creates an obtuse angle with the positive part of the axix $z$). Now, I have the function: $$ f\left(x,y,z\right)\:=\:ze^{x^2-y^2}-z$$ and I need to find the directional derivative of $f$ in the point in direction of $n$.
So the steps are easy: 
 1. Find the cruve $C$ as $\left(something\:\frac{,z}{2}\:,z\right)$
 2. Find the diravative of $C$ in the point.
 3. Find the gradient of $f$ in the point and do dot product with what I got in second step. 
My problems with this: 
 1. I dont know how to decide which equation to take when I find x as x(z) and how to use the information about the angle, ill be glad if someone explain it to me. 
 2. I tried to take the minus sqrt and I didnt get the result.

You have to take the positive square root since your point is $(1,1,2)$ where $x$ value is positive. 
Based on the information of the angle, the direction of the tangent has a negative $z$ value. So once you find the tangent direction, change the signs if necessary. 
