Let $\langle \cdot,\cdot \rangle$ an arbitrary inner product in $\mathbb{R}^n$ and $\|\cdot\|$ the norm induced by this inner product. Consider the function $f:\mathbb{R}^n\setminus\{0\}\to\mathbb{R}^n\setminus\{0\}$ given by $f(x)=\dfrac{x}{\|x\|^2}$.
I need to show that $f$ is a continuous bijection with continuous inverse.
First of all, I showed that $f$ is injective.
Consider $x,y\in\mathbb{R}^n$ such that $f(x)=f(y)$, then $\dfrac{x}{||x||^2}=\dfrac{y}{||y||^2}$. \begin{eqnarray*} \dfrac{x}{||x||^2}=\dfrac{y}{||y||^2} &\Rightarrow& \left\|\dfrac{x}{||x||^2} \right\|=\left\|\dfrac{y}{||y||^2} \right\|\\ &\Rightarrow& \dfrac{||x||}{||x||^2}=\dfrac{||y||}{||y||^2} \\ &\Rightarrow& \dfrac{1}{||x||}=\dfrac{1}{||y||} \\ &\Rightarrow& ||x||=||y||\\ \end{eqnarray*} Therefore, $\dfrac{x}{||x||^2}=\dfrac{y}{||x||^2} \Rightarrow x=y$.
Now, to show that $f$ is continuous. Consider $x_k\to x$, $x\neq 0$, then $||x_k||\to ||x||$ by continuity of norm function. Therefore, $$\dfrac{x^k}{||x^k||^2}\to\dfrac{x}{||x||^2},$$ i.e, $$f(x^k)\to f(x).$$ Concluding that $f$ is a continuous function.
Note that $f^{-1}(x)=\dfrac{x}{||x||^2}$. In fact, let $y=\dfrac{x}{||x||^2}$, the $$ ||y||=\dfrac{||x||}{||x||^2}=\dfrac{1}{||x||} \Rightarrow ||x||=\dfrac{1}{||y||}$$ Therefore, we have that $x=y||x||^2=\dfrac{y}{||y||^2}$. Thus the continuity of $f^{-1}$ follow by continuity of $f$.
However, I didn't get to show that $f$ is a subjective function.
If you've found a formula for the inverse function of $f$ (although you don't yet know it's defined on the entire codomain), then given $y\neq0$ it would be 'natural' to try find its source using this formula. Indeed, using your formula:
$$f^{-1}(y)=\frac{y}{||y||^2}=t\in\mathbb R^n/\{0\}$$ And you can check (just as you did before) that: $$f(t)=\frac{t}{||t||^2}=\frac{\frac{y}{||y||^2}}{\frac{1}{||y||^2}}=y$$ And so $f$ is surjective.