It is often easier to simplify first, and then differentiate. Let's see if we can rewrite your product of fractions:
[tex]z^2\cdot \frac{-(z-\frac{1}{z})^2}{20+8(z+\frac{1}{z})} \cdot \frac{1}{iz}<br />
= \frac{z^2[-(z^2-2+\frac{1}{z^2})]}{20+8(z+\frac{1}{z})} \cdot \frac{1}{iz}<br />
= \frac{-(z^4 + 2z^2 + 1)}{[20+8(z+\frac{1}{z})]z} \cdot \frac{1}{i}<br />
= -\frac{z^4 + 2z^2 + 1}{20z+8z^2+8} \cdot \frac{1}{i}<br />
= -\frac{z^4+2z+1}{8z^2+20z+8} \cdot (-i)[/tex]
which finally simplifies to
[tex]i \cdot \frac{z^4 + 2z + 1}{8z^2+20z+8} =<br />
\frac{i}{4} \cdot \frac{z^4+2z^2+1}{2z^2+5z+2}[/tex]
You can stop here and use the quotient rule for the first derivative; then differentiate the result to obtain the second derivative.
You may find it easier to write this fraction as a product (I personally consider the product rule to be cleaner and faster than the quotient rule, though they are equivalent in many respects).[tex]
\frac{i}{4} \cdot \frac{z^4+2z^2+1}{2z^2+5z+2} =<br />
\frac{i}{4} (z^4+2z+1) \cdot (2z^2+5z+2)^{-1}[/tex]
Using this form, just differentiate using the product rule, and don't simplify the result (because it makes finding the second derivative easier; we just apply the product rule again).