Solutions

(i) Proof by cases/exhaustion:
$\bm{x, y \geq 0}$

$$\begin{aligned} [ x \geq 0 \wedge y \geq 0 \implies x = |x| \wedge y = |y| ] \implies |xy| = xy = |x| \cdot |y| \end{aligned}$$

$\bm{x, y \leq 0}$

$$\begin{aligned} [ x \leq 0 \wedge y \leq 0 \implies -x = |x| \wedge -y = |y| ] \implies |xy| = xy = -x \cdot -y = |x| \cdot |y| \end{aligned}$$

$\bm{x \geq 0, y \leq 0}$

$$\begin{aligned} [ x \geq 0 \wedge y \leq 0 \implies x = |x| \wedge -y = |y| ] \implies |xy| = x \cdot -y = |x| \cdot |y| \end{aligned}$$

We can ignore the case $x \leq 0, y \geq 0$ as it would be treated
by the third one, if the starting inequalities' variables were
interchanged.

(ii)

$$\begin{aligned} |\frac{1}{x}| & = |1 \cdot x^{-1}| \\ & = |1| \cdot |x^{-1}| && (i) \\ & = 1 \cdot |x^{-1}| \\ & = \frac{1}{|x|} \end{aligned}$$

(iii)

$$\begin{aligned} \frac{|x|}{|y|} & = |x| |y|^{-1} \\ & = |x| |y^{-1}| && (ii) \\ & = |\frac{x}{y}| && (i) \end{aligned}$$

(iv)

$$\begin{aligned} |x + z| \leq |x| + |z| & \implies |x + z| \leq |x| + |-z| && z = -y \\ & \implies |x + (-y)| \leq |x| + |y| \\ & \implies |x - y| \leq |x| + |y| \end{aligned}$$

(v)

$$\begin{aligned} |x| = |(x - y) + y| & \leq |x - y| + |y| && \text{Triangle inequality} \\ \implies |x| - |y| & \leq |x - y| \end{aligned}$$

(vi)

$$\begin{aligned} |x| - |y| \leq |x - y| & \implies |y| - |x| \leq |y - x| && \text{Interchanging } x, y \\ & \implies |y| - |x| \leq |-(x - y)| \\ & \implies |y| - |x| \leq |x - y| \\ & \implies -|x - y| \ leq |x| - |y| \\ & \implies -|x - y| \leq |x| - |y| \leq |x - y| && (v) \\ & \implies ||x| - |y|| \leq |x - y| \end{aligned}$$

(vii)

$$\begin{aligned} |x + y| \leq |x| + |y| \implies |x + y| + |z| & \leq |x| + |y| + |z| && \text{Triangle inequality} \\ \implies |(x + y) + z| \leq |x + y| + |z| & \leq |x| + |y| + |z| \end{aligned}$$