Derivative of $$$\frac{z_{0}^{4}}{1 - z_{0}}$$$

Apply the quotient rule $$$\frac{d}{dz_{0}} \left(\frac{f{\left(z_{0} \right)}}{g{\left(z_{0} \right)}}\right) = \frac{\frac{d}{dz_{0}} \left(f{\left(z_{0} \right)}\right) g{\left(z_{0} \right)} - f{\left(z_{0} \right)} \frac{d}{dz_{0}} \left(g{\left(z_{0} \right)}\right)}{g^{2}{\left(z_{0} \right)}}$$$ with $$$f{\left(z_{0} \right)} = z_{0}^{4}$$$ and $$$g{\left(z_{0} \right)} = 1 - z_{0}$$$:

$${\color{red}\left(\frac{d}{dz_{0}} \left(\frac{z_{0}^{4}}{1 - z_{0}}\right)\right)} = {\color{red}\left(\frac{\frac{d}{dz_{0}} \left(z_{0}^{4}\right) \left(1 - z_{0}\right) - z_{0}^{4} \frac{d}{dz_{0}} \left(1 - z_{0}\right)}{\left(1 - z_{0}\right)^{2}}\right)}$$

The derivative of a sum/difference is the sum/difference of derivatives:

$$\frac{- z_{0}^{4} {\color{red}\left(\frac{d}{dz_{0}} \left(1 - z_{0}\right)\right)} + \left(1 - z_{0}\right) \frac{d}{dz_{0}} \left(z_{0}^{4}\right)}{\left(1 - z_{0}\right)^{2}} = \frac{- z_{0}^{4} {\color{red}\left(\frac{d}{dz_{0}} \left(1\right) - \frac{d}{dz_{0}} \left(z_{0}\right)\right)} + \left(1 - z_{0}\right) \frac{d}{dz_{0}} \left(z_{0}^{4}\right)}{\left(1 - z_{0}\right)^{2}}$$

The derivative of a constant is $$$0$$$:

$$\frac{- z_{0}^{4} \left({\color{red}\left(\frac{d}{dz_{0}} \left(1\right)\right)} - \frac{d}{dz_{0}} \left(z_{0}\right)\right) + \left(1 - z_{0}\right) \frac{d}{dz_{0}} \left(z_{0}^{4}\right)}{\left(1 - z_{0}\right)^{2}} = \frac{- z_{0}^{4} \left({\color{red}\left(0\right)} - \frac{d}{dz_{0}} \left(z_{0}\right)\right) + \left(1 - z_{0}\right) \frac{d}{dz_{0}} \left(z_{0}^{4}\right)}{\left(1 - z_{0}\right)^{2}}$$

Apply the power rule $$$\frac{d}{dz_{0}} \left(z_{0}^{n}\right) = n z_{0}^{n - 1}$$$ with $$$n = 1$$$, in other words, $$$\frac{d}{dz_{0}} \left(z_{0}\right) = 1$$$:

$$\frac{z_{0}^{4} {\color{red}\left(\frac{d}{dz_{0}} \left(z_{0}\right)\right)} + \left(1 - z_{0}\right) \frac{d}{dz_{0}} \left(z_{0}^{4}\right)}{\left(1 - z_{0}\right)^{2}} = \frac{z_{0}^{4} {\color{red}\left(1\right)} + \left(1 - z_{0}\right) \frac{d}{dz_{0}} \left(z_{0}^{4}\right)}{\left(1 - z_{0}\right)^{2}}$$

Apply the power rule $$$\frac{d}{dz_{0}} \left(z_{0}^{n}\right) = n z_{0}^{n - 1}$$$ with $$$n = 4$$$:

$$\frac{z_{0}^{4} + \left(1 - z_{0}\right) {\color{red}\left(\frac{d}{dz_{0}} \left(z_{0}^{4}\right)\right)}}{\left(1 - z_{0}\right)^{2}} = \frac{z_{0}^{4} + \left(1 - z_{0}\right) {\color{red}\left(4 z_{0}^{3}\right)}}{\left(1 - z_{0}\right)^{2}}$$

Simplify:

$$\frac{z_{0}^{4} + 4 z_{0}^{3} \left(1 - z_{0}\right)}{\left(1 - z_{0}\right)^{2}} = \frac{z_{0}^{3} \left(4 - 3 z_{0}\right)}{\left(z_{0} - 1\right)^{2}}$$

Thus, $$$\frac{d}{dz_{0}} \left(\frac{z_{0}^{4}}{1 - z_{0}}\right) = \frac{z_{0}^{3} \left(4 - 3 z_{0}\right)}{\left(z_{0} - 1\right)^{2}}$$$.