Note: Steepest ascent is from google translate, let me know if it makes no sense or there's another word to call it (I mean when the directional derivative is on the direction of the gradient vector)
Let $S$ be a surface with the function $z=f(x,y)=10-2x^2-y^2$, from a point starts a curve $L$ that is given by the parametric presentation: $x(t)=e^{-4t}$, $y(t)=2e^{-2t}$, $z(t)=10-2e^{-8t}-4e^{-4t}$.
Prove that in every point on the curve, the vector that is pointing in the direction of the steepest ascent in every point $M$ that is traveling on $L$ is : $(f_x(x,y),f_y(x,y),|\nabla f(x,y)|^2)$, what is your conclusion?
I know how to proof this for a specific point $M(x_0,y_0,f(x_0,y_0))$, by finding the gradient vector using the dot product with a vector that is vertical to the gradient vector in a specific point $M$.
I'm not sure though how to do it in general, the only approach I have is building $f(x(t),y(t))$, but also I'm not sure how to continue, I would be needing to write a general gradient vector for every point on $L$, I'm not sure how to move on with so many generalizations and feeling a little lost.
About the conclusion, I know that it must be that in every point $M$ the direction of the steepest ascent is the direction of the gradient vector.
Any tips or hints and feedback about my thinking and logic would be appreciated, thanks in advance.