If I have a function such as $z=(x^a + y^b)^2$ with $a$ and $b$ both greater than one... is it enough to show that it is not quasiconcave by showing that the second derivatives are not negative?
The reason I ask this is because here: http://www.econ.ucsb.edu/~tedb/Courses/GraduateTheoryUCSB/answersmid11.pdf
It says that $x^2$ on the positive real numbers is quasiconcave. However, the second derivative of $x^2$ is positive.
The quasi- part matters. A quasiconcave function need not have negative second derivative. It seems that the definition of quasiconcavity that you are using is $$f(ta+(1-t)b) \ge \min(f(a),f(b)),\quad 0\le t\le 1$$ It is a bit more convenient to work with an equivalent form of the above: for every $\lambda$, the set $\{x:f(x) \ge \lambda\}$ is convex (on a line this means that the set is an interval). For example, every increasing function on $[0,\infty)$ is quasiconcave, and $x^2$ is an example of that.
Concerning $f(x,y)=(x^a + y^b)^2$: note that the upper level set $\{(x,y): f(x,y)\ge \lambda\}$ can be written as $\{(x,y): x^a+y^b \ge \lambda^{1/2}\}$. To show this set is not always convex, I would try $\lambda=1$ and the points $(1,0)$ and $(0,1)$. The midpoint $(1/2,1/2)$ is not in the upper level set.