Given $f(x):\mathbb{R}^n\rightarrow\mathbb{R}$, if for any curve $\gamma:[-\epsilon,\epsilon]\rightarrow\mathbb{R}^n:t\mapsto(\gamma_1(t),\ldots,\gamma_n(t))$ such that $\gamma(0)=0$ and each $\gamma_i$ is real-analytic, we have that $f\circ\gamma(t)$ is real-analytic, can we say that $f$ is real-analytic at 0? If not, can we at least say that $f$ is smooth at $0$?
My thoughts are as follows. We want to construct a series such that $f(x)=\sum_{\alpha}c_\alpha x^\alpha$ where the sum is over all finite vectors $\alpha$ in $\mathbb{N}^m$ for some $m$ and $x^{\alpha}=x_1^{\alpha_1}\ldots x_m^{\alpha_m}$. Because $f\circ\gamma$ is analytic for linear $\gamma$ we can find constants $c_\alpha$ for $\alpha=(i,i,\ldots,i)^m$ for some $i,\,m$ such that $f(x)=\sum_{\alpha}c_\alpha x^\alpha$ for $x$ on the axes in $\mathbb{R}^n$ sufficiently close to $0$. Similarly, for any $x_0\in\mathbb{R}^n\setminus \{0\}$ $f(tx_0)=\sum_{n=0}^\infty d_n(x_0)t^n$ for some sequence of constants $d_n(x_0)$ and $t$ sufficiently small, and there are (big claim) a unique set of constants $c_\alpha$ such that $\sum_{n=0}^\infty d_n(x_0)t^n=\sum_{\alpha}c_{\alpha}(tx_0)^\alpha$ for all choices of $x_0$, at least formally. As the infimum of the radius of convergence (in $t$) of $f\circ \gamma$ over linear $\gamma$ is bounded away from $0$ by compactness of the unit sphere we have convergence of $\sum_{\alpha}c_{\alpha}x^\alpha$ in an open neighbourhood of $0$.
I found a solution. As smoothness along any smooth curve implies smoothness by Boman's Theorem we have a power series $\sum_{\alpha}c_{\alpha}x^\alpha$ for $f$. As $f(tx_0)$ is analytic for any $x_0$ in the unit sphere the power series $\sum_{\alpha}c_{\alpha}t^{|\alpha|}x_0^\alpha$ converges for $|t|<R_{x_0}$ and is equal to $f(tx_0)$. As $R=\inf_{|x_0|=1}R_{x_0}>0$ by compactness, for any $|x|<R$ we have $x=tx_0$ for some $|x_0|=1$, $t<R\leq R_{x_0}$ and so $f(x)=f(tx_0)=\sum_{\alpha}c_{\alpha}t^{|\alpha|}x_0^\alpha=\sum_{\alpha}c_{\alpha}x^{\alpha}$.