Differentiability along hyperplanes

Question

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.

Answer 1

No, $f$ need not be continuous at $0$, here is a counterexample.

Let $(A_n)_n$ be a sequence of balls $A_n=B(p_n,r_n)$ with $0\not\in A_n$, $p_n\to0$, $r_n\to0$ and such that, for any $n_1<n_2<n_3\in\mathbb{N}$ and any vectors $v_i\in A_{n_i}$, the vectors $v_1,v_2,v_3$ are linearly independent (see a construction below).

And let $\phi_n$ be a smooth bump function supported inside $A_n$ with $\|\phi_n\|_\infty=1$. Then $f=\sum_n\phi_n$ is not continuous at $0$, but its restriction to any plane is a sum of two smooth bump functions, so it's smooth.

An example of a sequence of balls as above:

Let $p_n=\left(\frac{1}{10^n},\frac{1}{100^n},\frac{1}{1000^n}\right)$ and $r_n=\frac{1}{1000000^{n}}$, for $n=10,100,1000,\dots$

Now let $n_1<n_2<n_3\in\{10,100,1000,\dots\}$ and let $|\varepsilon_{i,j}|<\frac{1}{1000000^{n_i}}$, it will be enough to check that the following matrix is invertible:

$$\begin{pmatrix} \frac{1}{10^{n_1}}+\varepsilon_{1,1}&\frac{1}{100^{n_1}}+\varepsilon_{1,2}&\frac{1}{1000^{n_1}}+\varepsilon_{1,3}\\ \frac{1}{10^{n_2}}+\varepsilon_{2,1}&\frac{1}{100^{n_2}}+\varepsilon_{2,2}&\frac{1}{1000^{n_2}}+\varepsilon_{2,3}\\ \frac{1}{10^{n_3}}+\varepsilon_{3,1}&\frac{1}{100^{n_3}}+\varepsilon_{3,2}&\frac{1}{1000^{n_3}}+\varepsilon_{3,3}\\ \end{pmatrix}$$

This determinant of this matrix is a sum of $48$ terms, but the term $\frac{1}{10^{n_3}100^{n_2}1000^{n_1}}$ is bigger in absolute value than all other terms by a factor $>100$ (it's enough to check this for the other five terms not involving the $\varepsilon_{i,j}$), so the determinant is not $0$.