How to prove that these Jacobians are equal?
$$\dfrac{\partial (x,y)}{\partial(\alpha, \beta)} \cdot \dfrac{\partial(\alpha, \beta)}{\partial(z,w)} = \dfrac{\partial (x,y)}{\partial(z,w)}$$
I don't find it staightforward to cancel $\partial(\alpha, \beta)$ out, as it is operator but not a fraction. But I have tried to express them following the definition, without success though.
Thanks.
Use the chain rule. You have on the left-hand-side: $$\begin{bmatrix} \frac{\partial x}{\partial \alpha} & \frac{\partial x}{\partial \beta} \\ \frac{\partial y}{\partial \alpha} & \frac{\partial y}{\partial \beta} \end{bmatrix} \begin{bmatrix} \frac{\partial \alpha}{\partial z} & \frac{\partial \alpha}{\partial w} \\ \frac{\partial \beta}{\partial z} & \frac{\partial \beta}{\partial w}\end{bmatrix}.$$ The determinant of the product of two matrices is the product of the determinants. The first entry of the above product is, for example: $$\frac{\color{blue}{\partial x}}{\partial \alpha}\frac{\partial \alpha}{\color{blue}{\partial z}}+\frac{\color{blue}{\partial x}}{\partial \beta}\frac{\partial \beta}{\color{blue}{\partial z}}= \color{blue}{\frac{\partial x}{\partial z}},$$ which is the first entry of the matrix that you'll compute the determinant of the right-hand-side. Do the same for the other entries.
You can't just cancel $\partial(\alpha, \beta)$ right off the bat, Jacobians are not fractions. But this exercise tells you that for practical purposes, you can "cancel" $\partial(\alpha,\beta)$ indeed.