$X$ Cauchy distributed then $Y=\frac1X$ also Cauchy distributed

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If $X$ is Cauchy distributed then $Y:=\frac1X$ is also Cauchy distributed.

It holds:

$$y=g(x)=\frac{1}{x} \Rightarrow g^{-1}(y)=\frac{1}{y} \Rightarrow (g^{-1}(y))'= -\frac{1}{y^2} $$

How does then follow for density function:

$$f_Y(y)=f_X(g^{-1}(y)) \cdot | (g^{-1}(y))'|$$

Why do I need $|\cdot|$?

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When transforming random variables, the transformed variable still has to respect the probability axioms, in particular that probability cannot be negative. Hence the absolute value around $(g^{-1}(y))'$ is necessary, otherwise $f_Y(y)$ could assume negative values (and here it would, because $(g^{-1}(y))'=-\frac1{y^2}<0$).