Okay, so this is what I came up with:

Code:

t=-2:0.1:2;
x1=cosh(t);y1=sinh(t);x2=-cosh(t);y2=sinh(t);
dx1=2*sinh(t);dy1=-2*cosh(t);dx2=-2*sinh(t);dy2=dy1;
plot(x1,y1)
hold on
plot(x2,y2)
hold on
quiver(x1,y1,dx1,dy1)
hold on
quiver(x2,y2,dx2,dy2)
axis equal
xlabel('x');ylabel('y');zlabel('z')
grid on
title('Exercise 8.2 by Brent Daniels')

The result is this figure:

This doesn't seem right, though. $\displaystyle \nabla f(x,y) = <2x,-2y>$ since this figure is essentially a level curve of $\displaystyle z=x^{2}-y^{2}$. Therefore, shouldn't all the normal vectors on the lower halves of the two branches be pointing in the opposite direction? Or is my code/approach wrong? Thanks.

Edit: Nevermind, I'm an idiot and found my mistake. What essentially I was plotting was $\displaystyle <2\frac{dx}{dt},-2\frac{dy}{dt}>$ instead of $\displaystyle <2x,-2y>$. Thanks again.