We say that a function $f : \mathbb{R}^n \rightarrow \mathbb{R}^k$ is $n$-homogeneous if $f(\lambda \vec{x}) = \lambda^n f(\vec{x})$ for all $\vec{x}$ and $\lambda > 0$ where $n>0$.
How could I prove the continuity of an $n$-homogeneous $f$ at the origin?
I have considered trying to use $\epsilon-\delta$, but it didn't get me anywhere.
On
Such a function is not necessarily continuous at the origin. This is trivial unless we strengthen your definition to apply for $\lambda \geq 0$, not just $\lambda > 0$. In that case, we find a 2-homogeneous counterexample $f: \mathbb{R}^{2} \to \mathbb{R}$. First of all, we can find an unbounded (necessarily discontinous) function $g: S^{1} \to \mathbb{R}$ for $S^{1}$ the unit circle, since $S^{1}$ is infinite. Then we let $f(x)=||x||^{2}g(\frac{x}{||x||})$ for $x\in \mathbb{R}^{2} \backslash \{(0,0)\}$ and $f(0)=0$.
The assumption that $f$ is $n$-homegeneous implies $f(0)=0.$ To see this, take $x=0.$ We then have $f((1/2)\cdot 0)=f(0) = (1/2)^nf(0).$ That's not possible unless $f(0)=0.$
The following result answers your question: Assume $f$ is $n$-homogeneous. Then $f$ is continuous at $0$ iff $f$ is bounded on the unit sphere $S.$
Proof: $\implies$: Suppose to reach a contradiction that $f$ is not bounded on $S.$ Then for $k=1,2,\dots,$ there exist $u_k \in S$ such that $|f(u_k)|>2^{kn}.$ The sequence $(1/2^k)u_k \to 0,$ yet
$$|f((1/2^k)u_k)|=|(1/2^{kn})f(u_k)| >1,\,\, k=1,2,\dots$$
Thus $f$ fails to converge to $0=f(0)$ along a sequence converging to $0.$ But $f$ is continuous at $0,$ contradiction.
$\impliedby$: Let $M=\sup_S |f|.$ Let $\epsilon>0.$ Set $\delta =(\epsilon/M)^{1/n}.$ If $0<|x|< \delta,$ then
$$|f(x)-f(0)| = |f(x)|= |f(|x|(x/|x|))| \le |x|^nM < \delta^n M = \epsilon.$$