I am reading "Analysis on Manifolds" by James R. Munkres.
(1) Let $\alpha:\mathbb{R}\to\mathbb{R}^2$ be the mapping such that $\alpha(t) = (t^3,t^2)$ for $t\in\mathbb{R}$.
Then, $\alpha$ is one-to-one and $\alpha^{-1}:\alpha(\mathbb{R})\to\mathbb{R}$ is continuous.
My proof:
$t\mapsto t^3$ is one-to-one.
So, $\alpha$ is one-to-one.
Let $(u_0,v_0)\in\alpha(\mathbb{R})$.
Let $t_0\in\mathbb{R}$ be the unique real number such that $t_0^3=u_0$ and $t_0^2=v_0$.
Let $\varepsilon_0$ be an arbitrary positive real number.
We want to find a positive real number $\delta$ such that $$||(u,v)-(u_0,v_0)||<\delta\,\,\text{ and } (u,v)\in\alpha(\mathbb{R})\implies |\alpha^{-1}((u,v))-t_0|<\varepsilon_0.$$
Since $\alpha^{-1}((u,v))=u^{\frac{1}{3}}$ and $u\mapsto u^{\frac{1}{3}}$ is continuous, there is a positive real number $\delta_0$ such that $$|u-u_0|<\delta_0\implies |u^{\frac{1}{3}}-u_0^{\frac{1}{3}}|=|u^{\frac{1}{3}}-t_0|<\varepsilon_0.$$
And since $|u-u_0|\leq ||(u,v)-(u_0,v_0)||$, $$||(u,v)-(u_0,v_0)||<\delta_0\,\,\text{ and } (u,v)\in\alpha(\mathbb{R})\implies |\alpha^{-1}((u,v))-t_0|<\varepsilon_0$$ holds.
(2) Let $\alpha:\mathbb{R}^2\to\mathbb{R}^3$ be the mapping such that $\alpha(x,y) = (x(x^2+y^2),y(x^2+y^2),x^2+y^2)$ for $(x,y)\in\mathbb{R}^2$.
Then, $\alpha$ is one-to-one and $\alpha^{-1}:\alpha(\mathbb{R}^2)\to\mathbb{R}^2$ is continuous.
I tried to prove $\alpha^{-1}$ is continuous but I couldn't prove it.
My attempt:
Let $(x,y)\neq (x',y')$.
If $x^2+y^2\neq x'^2+y'^2$, then $$(x(x^2+y^2),y(x^2+y^2),x^2+y^2)\neq (x'(x'^2+y'^2),y'(x'^2+y'^2),x'^2+y'^2).$$
If $x^2+y^2=x'^2+y'^2$, then $x^2+y^2=x'^2+y'^2\neq 0$ because $(x,y)\neq (x',y')$.
So, if $x^2+y^2=x'^2+y'^2$, then $x(x^2+y^2)\neq x'(x'^2+y'^2)$ or $y(x^2+y^2)\neq y'(x'^2+y'^2)$ holds.
So, $\alpha^{-1}$ is one-to-one.
Let $\varepsilon$ be an arbitrary positive real number.
It is sufficient to prove there is a positive real number $\delta$ such that $$\sqrt{[x(x^2+y^2)-x_0(x_0^2+y_0^2)]^2+[y(x^2+y^2)-y_0(x_0^2+y_0^2)]^2+[(x^2+y^2)-(x_0^2+y_0^2)]^2}<\delta\implies\sqrt{(x-x_0)^2+(y-y_0)^2}<\varepsilon.$$
Since $\alpha(0,0)=(0,0,0)$, $\alpha^{-1}(0,0,0)=(0,0)$.
Let $(X,Y,Z)\in\alpha(\mathbb{R}^2)\setminus\{(0,0,0)\}$.
Then $X^2+Y^2=Z^3$.
So, $Z=(X^2+Y^2)^{\frac{1}{3}}$.
Then, $\alpha^{-1}(X,Y,Z)=\left(\frac{X}{(X^2+Y^2)^{\frac{1}{3}}},\frac{Y}{(X^2+Y^2)^{\frac{1}{3}}}\right)$ holds.
We check this:
$\alpha\left(\frac{X}{(X^2+Y^2)^{\frac{1}{3}}},\frac{Y}{(X^2+Y^2)^{\frac{1}{3}}}\right)=\left(\frac{X}{(X^2+Y^2)^{\frac{1}{3}}}\cdot(X^2+Y^2)^{\frac{1}{3}},\frac{Y}{(X^2+Y^2)^{\frac{1}{3}}}\cdot(X^2+Y^2)^{\frac{1}{3}},(X^2+Y^2)^{\frac{1}{3}}\right)=(X,Y,Z)$.
Let $\beta:\mathbb{R}^2\to\mathbb{R}^2$ be the mapping such that $\beta(0,0)=(0,0)$ and $\beta(X,Y)=\left(\frac{X}{(X^2+Y^2)^{\frac{1}{3}}},\frac{Y}{(X^2+Y^2)^{\frac{1}{3}}}\right)$ for $(X,Y)\neq (0,0)$.
Then, $\beta$ is continuous at $(X,Y)\neq (0,0)$.
We check $\beta$ is continuous at $(0,0)$.
It is sufficient to check the following:
For an arbitrary positive real number $\varepsilon$, there is a positive real number $\delta$ such that $0<\sqrt{X^2+Y^2}<\delta\implies\sqrt{\frac{X^2}{(X^2+Y^2)^{\frac{2}{3}}}+\frac{Y^2}{(X^2+Y^2)^{\frac{2}{3}}}}<\varepsilon$.
Let $\delta:=\varepsilon^3$.
Then the above holds.
$\alpha^{-1}$ is continuous at $(X,Y,Z)\neq (0,0,0)$.
We check $\alpha^{-1}$ is continuous at $(0,0,0)$.
It is sufficient to check the following:
For an arbitrary positive real number $\varepsilon$, there is a positive real number $\delta$ such that $(X,Y,Z)\in\alpha(\mathbb{R}^2)$ and $0<\sqrt{X^2+Y^2+Z^2}<\delta\implies\sqrt{\frac{X^2}{(X^2+Y^2)^{\frac{2}{3}}}+\frac{Y^2}{(X^2+Y^2)^{\frac{2}{3}}}}<\varepsilon$.
But this obviously holds because $\sqrt{X^2+Y^2}\leq\sqrt{X^2+Y^2+Z^2}$.
But I don't know how to prove the inverse function $\alpha^{-1}$ is continuous in general.