I read the wiki page on PDF and I got stuck at the proof of relation between pdf $g$ of a function $\textbf{y}$ and pdf $f$ of its random variables $\textbf{x}$:
$$g({\bf{y}}) = f({H^{ - 1}}({\bf{y}}))\left| {\det\left[ {{{\left. {\frac{{d{H^{ - 1}}({\bf{z}})}}{{d{\bf{z}}}}} \right|}_{{\bf{z}} = {\bf{y}}}}} \right]} \right|,$$ where $\textbf{y} = H (\textbf{x})$, $f$ - pdf for $\bf{x}$
This is proved from the equation $$\left|g(\textbf{y})\,d\textbf{y}\right|=\left|f(\textbf{x})\,d\textbf{x}\right|.$$ Since the probability is always positive we can rewrite it like $$g(\textbf{y})\left|\,d\textbf{y}\right|=f(\textbf{x})\left|\,d\textbf{x}\right|.$$ Now we have two numbers that can be expanded as follows: $$\left|\,d\textbf{y}\right| = \sqrt{{dy_1}^2 + \cdots + {dy_n}^2}$$ $$\left|\,d\textbf{x}\right| = \sqrt{{dx_1}^2 + \cdots + {dx_n}^2}$$ Let's divide it $\frac{\left|\,d\textbf{x}\right|}{\left|\,d\textbf{y}\right|} = \sqrt{\frac{{dx_1}^2 + \cdots + {dx_n}^2}{{dy_1}^2 + \cdots + {dy_n}^2}}$.
So how can we from this stage get to the $\frac{\left|\,d\textbf{x}\right|}{\left|\,d\textbf{y}\right|}=\left|\frac{d\textbf{x}}{d\textbf{y}}\right|$?
UPD:This step is easy when $\dim(\textbf{x}) = \dim(\textbf{y})$: $$\frac{{\left| {{\mkern 1mu} d{\bf{x}}} \right|}}{{\left| {{\mkern 1mu} d{\bf{y}}} \right|}} = \frac{{\sqrt {d{{\bf{x}}^T}d{\bf{x}}} }}{{\sqrt {d{{\bf{y}}^T}d{\bf{y}}} }} = \sqrt {\frac{{d{{\bf{x}}^T}d{\bf{x}}}}{{d{{\bf{y}}^T}d{\bf{y}}}}} = \sqrt {\frac{{d{{\bf{x}}^T}}}{{d{{\bf{y}}^T}}}\frac{{d{\bf{x}}}}{{d{\bf{y}}}}} = \sqrt {{{\left( {\frac{{d{\bf{x}}}}{{d{\bf{y}}}}} \right)}^T}\frac{{d{\bf{x}}}}{{d{\bf{y}}}}} = \left| {\frac{{d{\bf{x}}}}{{d{\bf{y}}}}} \right|$$ But what to do if the dimensions equality is not the case? And how to show then that $\left| {\frac{{d{\bf{x}}}}{{d{\bf{y}}}}} \right| = {\rm{ }}\left| {\det\left( {\frac{{d{\bf{x}}}}{{d{\bf{y}}}}} \right)} \right|$?
See the answer to the 1st question in comment of @kimchi-lover The answer to the 2nd question: For 2D case the coordiantes transformation is [\begin{array}{*{20}{l}} {d{\bf{y_1}} = J({x_1},{x_2})d{{\bf{x}}_{\bf{1}}} = \left( {\frac{{d{y_1}}}{{d{x_1}}}d{x_1},\frac{{d{y_2}}}{{d{x_1}}}d{x_1},0} \right)}\\ {d\textbf{y}_2 = J({x_1},{x_2})d{{\bf{x}}_{\bf{2}}} = \left( {\frac{{d{y_1}}}{{d{x_2}}}d{x_2},\frac{{d{y_2}}}{{d{x_2}}}d{x_2},0} \right)}, \end{array} where ${d\bf{x}_1} = (dx_1,0)$, ${d\bf{x}_2} = (0, dx_2)$, ${d\bf{y}_1} = (dy_1,0)$ and ${d\bf{y}_2} = (0, dy_2)$. Then cross-product of $$d\textbf{y}_1 \times d\textbf{y}_2 = \left\langle {0,0,\left| {\begin{array}{*{20}{c}} {\frac{{d{y_1}}}{{d{x_1}}}}&{\frac{{d{y_1}}}{{d{x_2}}}}\\ {\frac{{d{y_2}}}{{d{x_1}}}}&{\frac{{d{y_2}}}{{d{x_2}}}} \end{array}} \right|} \right\rangle d{x_1}d{x_2} $$
Then the area in new coordinate system is $$dA = \left| \left| d\textbf{y}_1 \times d\textbf{y}_2 \right| \right| = \left| {\frac{{d{y_1}}}{{d{x_1}}}\frac{{d{y_2}}}{{d{x_2}}} - \frac{{d{y_1}}}{{d{x_2}}}\frac{{d{y_2}}}{{d{x_1}}}} \right|d{x_1}d{x_2} $$
and finally $$dA = \left| d\textbf{y} \right| = \left| {\frac{{\partial ({y_1},{y_2})}}{{\partial ({x_1},{x_2})}}} \right|d{x_1}d{x_2} = \left| {\frac{{\partial ({y_1},{y_2})}}{{\partial ({x_1},{x_2})}}} \right|\left| {d\textbf{x}} \right|$$ and $$\left| {\frac{{d\textbf{y}}}{{d\textbf{x}}}} \right| = \left| {\frac{{\partial ({y_1},{y_2})}}{{\partial ({x_1},{x_2})}}} \right| = \left| {\det \left[ {\frac{{d\textbf{y}}}{{d\textbf{x}}}} \right]} \right|$$
Note: I've mistaken $\textbf{y}$ for $\textbf{x}$, but the idea should be clear anyway
UPD: What to do if $\left| \left| d\textbf{y}_1 \times d\textbf{y}_2 \right| \right| \neq \left| d\textbf{y} \right|$? Are the last 2 equations valid?