Difficulty understanding the gradient vector Please take a look here:

Assume I want to decide the sign of $\frac{dz}{dt}|_{t=1} $ (and not t=2). 
As far as I can see:
At $t=1$ i think that the derivative is positive... this is because the $x'(1)=0 ,y'(1)>0$ and $  z_y(3,0)  > 0 $ . In one of my previous posts, Babak S. said he thinks I'm right.
On the other hand, at the point $(3,0) $ , the gradient must be perpendicular to the corresponding level set, so it must point towards the positive region of the x-axis, which means that the y'th-coordinate term of the gradient vector must be $0$ and thus $\frac{dz}{dt}|_{t=1} = 0$ ... 
Can someone please explain to me what am I getting wrong here? 

Why do you say $z_y(3,0) > 0$? As you observed correctly in your second paragraph, the gradient is in the $x$-direction at $(3,0)$, so $z_y(3,0) = 0$.
