This is a follow up to my previous question.

Let $f\colon \mathbb R^3\to \mathbb R$ be a rational function that is differentiable along all planes through $0$, that is, we assume:

1. There are polynomials $p,q$ of three real variables such that $f(x)=p(x)/q(x)$.
2. For all linear maps $\varphi\colon \mathbb R^2\to \mathbb R^3$ the composition $f\circ \varphi$ is (totally) differentiable at $0\in \mathbb R^2$.

Question. Is $f$ necessarily (totally) differentiable at $0$?

• As answer to my previous question (which did not require $f$ to be rational), Saúl RM constructed a function $f\in C^\infty(\mathbb R^3\backslash 0)\cap C(\mathbb R^3)$ with $f\circ \varphi\in C^\infty(\mathbb R^2$) for all $\varphi$, but which is not differentiable at $0$.
• For homogeneous polynomials one should look for examples with $\deg(p)\ge \deg (q)+1$. If $\deg(p)<\deg(q)$, then $f$ will blow up at $0$ (and hence cannot be differentiable along planes) and for $\deg(p)=\deg(q)$ the function $f$ would be $0$-homogeneous and thus continuity at zero would already imply that it is constant.
• Instead of asking $f\circ \varphi$ to be (totally) differentiable, one could ask it to be a polynomial, which is to say that $q\circ \varphi$ divides $p\circ \varphi$ in the polynomial ring of $2$ variables. I have asked a related question here.

This is a follow up to my previous question.

Let $f\colon \mathbb R^3\to \mathbb R$ be a continuous function that is rational and differentiable along all planes through $0$, that is, we assume:

1. There are polynomials $p,q$ of three real variables such that $f(x)=p(x)/q(x)$.
2. For all linear maps $\varphi\colon \mathbb R^2\to \mathbb R^3$ the composition $f\circ \varphi$ is (totally) differentiable at $0\in \mathbb R^2$.

Question. Is $f$ necessarily (totally) differentiable at $0$?

• As answer to my previous question (which did not require $f$ to be rational), Saúl RM constructed a function $f\in C^\infty(\mathbb R^3\backslash 0)\cap C(\mathbb R^3)$ with $f\circ \varphi\in C^\infty(\mathbb R^2$) for all $\varphi$, but which is not differentiable at $0$.
• For homogeneous polynomials one should look for examples with $\deg(p)\ge \deg (q)+1$. If $\deg(p)<\deg(q)$, then $f$ will blow up at $0$ (and hence cannot be differentiable along planes) and for $\deg(p)=\deg(q)$ the function $f$ would be $0$-homogeneous and thus continuity at zero would already imply that it is constant.
• Instead of asking $f\circ \varphi$ to be (totally) differentiable, one could ask it to be a polynomial, which is to say that $q\circ \varphi$ divides $p\circ \varphi$ in the polynomial ring of $2$ variables. I have asked a related question here.