A homogenous polynomial of degree $k$. $f: \mathbb R^n \to \mathbb R$ is a smooth map such that $f(rv)=r^kf(v)$ for all $v\in \mathbb R^n$ and $r\in\mathbb R$. So $f$ has to be a homogenous polynomial of degree $k$? This is an assertion from _Differential Geometry_ by Taubes.

Let $r \in R \setminus \\{0\\}$ and $x \in R^n$. Therefore, $f(rx)= r^k f(x),$ now by taking derivative from both sides w.r.t $x$, we get $rf'(rx)= r^k f'(x)$ thus $f'(rx)= r^{k-1} f'(x)$ now if we take derivative $k$ times w.r.t $x$ we get $$f^{k}(rx)= f^{k}(x)$$. By letting $r \rightarrow 0$ we get $f^{k}(0)= f^{k}(x)$ for all $x \in R^n$. Again by taking derivate we get $f^{n}(x)= 0$ for all $n > k$ Using taylor expansion we may rewrite $f$ as $$f(x)= P(x) + f^{k+1}(y) \times \\{\text{a bunch of vectors}\\} =P(x) $$ where $P$ is a polynomial of degree at most $K$. Now if $f \
eq 0$ taking into account that $P(rx)= r^k P(x)$, one can easily conclude that $P$ has to be a homogenous polynomial of degree $k$