Steepest ascent and descent of a function For the function $z = x^2 + y^2/4$ , find the direction of its steepest descent,steepest ascent, and no change at the point $( \sqrt2/2, \sqrt2 )$ on the level curve `z = 1`. I think I'm supposed to compute the directional derivative. I have the point required to do this, but not a vector (the question's answer should be a vector). What am I missing?

The direction of steepest ascent is the gradient. The direction of steepest descent is the negative of the gradient. If you traveled along the ellipse $x^2 + y^2 / 4$ then there would be no change; this would be the tangent vector to the ellipse with z remaining constant at 1.