**Definition.** Let us say that a function $f\colon \mathbb R^d\to \mathbb R$ is *differentiable along hyperplanes in the point $0\in \mathbb R^d$*, if $f\circ \varphi\colon \mathbb R^{d-1}\to \mathbb R$ is (totally) differentiable in $0\in \mathbb{R}^{d-1}$ for all linear maps $\varphi\colon \mathbb R^{d-1}\to \mathbb R^d$.

For $d=2$ there are some standard counterexamples to show that this notion is weaker than total differentiablity, e.g.: $$ f(x,y)=\frac{xy^2}{x^2+y^2} $$ This is of the form $f=p/q$ for homogeneous polynomials with $\deg p = 1+\deg q$. Differentiability along lines follows from the fact that homogeneous polyomials in one variable are automatically divisible, if the degree of the enumerator is larger than that of the denominator.

**Question.** Suppose $f\colon \mathbb R^3\to \mathbb R$ is differentiable along planes in the point $0$. Is $f$ then also totally differentiable in $0$?

If we restrict to homogeneous rational functions $f=p/q$ with $\deg p \ge 1+\deg q$ one can also ask for the stronger property that $q\circ \varphi$ shall divide $p\circ \varphi$ in the polynomial ring of $2$ variables and wonder whether this already implies that $q$ divides $p$. I've asked this algebraic question on math.stackexchange but I haven't received a satisfying answer (it seems tricky that we deal with not necessarily reducible polynomials over $\mathbb R$). Even if settled, this doesn't answer the full question asked here.