Small changes $$\delta x$$ and $$\delta y$$ in the x,y co-ordinates will bring about a small change $$\delta z$$ in the z co-ordinate given by
$$\delta z \approx \frac{\partial z}{\partial x}\delta x+ \frac{\partial z}{\partial y}\delta y.$$
So, $$\delta z \approx 4x \delta x - 6y \delta y.$$
We need $$\delta z$$ to be negative, in which case we need $$\delta x$$ to be negative and $$\delta y$$ to be positive. So, the move needs to be into the second quadrant of the xy plane.
If then we move a small distance $$h$$ at an angle $$\theta$$ to the negative x-axis, $$\delta x = -h\cos \theta,$$ and $$\delta y = h \sin \theta,$$ so $$\delta z \approx -h(8\cos\theta+6\sin\theta).$$
This will be a maximum when $$\tan \theta = 3/4$$ from which it follows that a vector in the required direction will be $$-4\hat{\imath}+3\hat{\jmath}$$ and a unit vector will be that divided by 5.
Hello, im new
