ODEs: Lipschitz-continuity and an IVP

$\textbf{1)}$ Let $g, h$ be the identity function. Clearly $g$ is Lipschitz continuous and $h$ is continuous. Then $f$ maps $$\begin{bmatrix} y^1 \\ y^2 \end{bmatrix} \longmapsto \begin{bmatrix} y^1 \\ y^1y^2\end{bmatrix}.$$ Now for any $r \in \mathbb{R}$, $$\lVert f(r, s) - f(r, t) \rVert = |r(s-t)| = |r| \cdot \lVert(r,s) - (r,t)\rVert.$$ Thus no Lipschitz constant exists for $f$; hence $f$ is not Lipschitz continuous.

$\textbf{2)}$ Since $g$ is Lipschitz continuous, there is some open interval $J \subseteq \mathbb{R}$ containing $x_0$ on which there exists a unique solution $y_1 \colon J \to \mathbb{R}$ to the differential equation $$(y^1)'(x) = g(y^1(x)) \tag{$\star$}$$ with initial condition $y^1(x_0) = y^1_0$. Given $y^1$, define $y^2 \colon J \to \mathbb{R}$ by $$y^2(x) = y^2_0\exp\left(\int_{x_0}^{x} h(y^1(x)) \, \mathrm{d}x \right).$$ Now let $y \colon J \to \mathbb{R}^2$ be defined by $y(x) = (y^1(x), y^2(x))$. It follows that $$y'(x) = \begin{bmatrix} (y^1)'(x) \\ (y^2)'(x) \end{bmatrix} = \begin{bmatrix} g(y^1(x)) \\ h(y^1(x))y^2(x) \end{bmatrix}.$$ Furthermore, $y^1(x_0) = y^1_0$ and $y^2(x_0) = y^2_0$ by construction. Hence $y$ is a solution to the original IVP. To show uniqueness, suppose $z \colon J \to \mathbb{R}$ is also a solution to the IVP. Then $(z^1)' = g(z^1)$ on $J$ and $z^1(x_0) = y^1_0$, which by uniqueness of the solution to $(\star)$ implies that $z^1 = y^1$ on $J$. Furthermore, one has $$(z^2)' = h(z^1)z^2 \implies \frac{(z^2)'}{z^2} = h(z^1)$$ The left hand side is the logarithmic derivative, that is $$\frac{\rm d}{\mathrm{d}x} \log(z^2(x)) = h(z^1(x)) \implies \log(z^2(x)) = \log(z^2_0) + \int_{x_0}^{x} h(z^1(x)) \, \mathrm{d} x.$$ Exponentiating and using the fact that $z^1 = y^1$ on $J$ yields that $z^2 = y^2$ on $J$. Thus $y$ is the unique solution to the IVP on $J$.