Understanding the partial derivative of temperature with respect to thermodynamic beta (coldness)
You know that:


  $$\beta = \frac{1}{k_\mathrm BT}$$


Rearranging yields:
$$\color{red}{\beta} = \frac{1}{k_\mathrm B\color{blue}{T}}\implies \color{blue}{T} = \frac{1}{k_\mathrm B \color{red}{\beta}}$$

Thus:

$$\frac{\partial \color{blue}{T}}{\partial\beta} = \frac{\partial}{\partial\beta}\left(\frac{1}{k_\mathrm B \beta}\right) = \color{red}{-}\frac{1}{k_\mathrm B \beta^\color{red}{2}}$$

Substituting $\beta = \color{red}{\frac{1}{k_\mathrm BT}}$ yields:

$$\frac{\partial T}{\partial\beta} =  -\frac{1}{k_\mathrm B \color{red}{\left(\frac{1}{k_\mathrm BT}\right)}^2} = -\frac{k_\mathrm B^2T^2}{k_\mathrm B} = -k_\mathrm BT^2$$

$$-\frac{\partial T}{\partial\beta} = k_\mathrm BT^2$$
Q.E.D.