calculus - One-variable continuity of one partial derivative implies differentiability?

I am interested in a particular instacne of the phenomena

"Partial derivatives + (A certain degree of) continuity" implies differentiablilty. My case assumes less regularity than usual:

Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}, (x_0,y_0) \in \mathbb{R}^2$.

Assume that the partial derivatives of $f$ with respect to $x$ and $y$ exist at the point $(x_0,y_0)$ and one of them exists and continuous w.r.t to the other variable (for instance the function $y \mapsto\frac{\partial{f}}{\partial x}(x_0,y)$ is continuous at the point $y_0$).

Is it true that $f$ is differentiable at $(x_0,y_0)$?

Note: It is known that if both partial derivatives exists, and one of them is continuous (as a functions of two variables) then $f$ is differentiable. (For a proof see here).

However, the proof uses:

1) The existence of $\frac{\partial{f}}{\partial x}$ on some ball around $(x_0,y_0)$

(I assume only $\frac{\partial{f}}{\partial x}$ exists on $\{x_0\} \times (y_0-\epsilon,y_0+\epsilon)$).

2) The continuity of $(x,y) \mapsto \frac{\partial{f}}{\partial x}(x,y)$ at $(0,0)$.

(I assume only continuity of $\frac{\partial{f}}{\partial x}(x_0,y)$).

Answer

This is false.

Counterexample, at the point $(x_0,y_0)=(0,0)$:
$$f(x,y) = \begin{cases} x^3/y, & y\neq 0 ,\\ 0, & y=0 . \end{cases}$$
This function satisfies the hypotheses:

• $f_y(0,0)=0$ (since $f$ is constant along the $y$ axis: $f(0,y)=0$ for all $y$),
  so in particular $f_y(0,0)$ exists.


• $f_x(0,y) = 0$ for all $y$ (since the restriction of $f$ to a line $y=c$ is either identically zero or a constant times $x^3$, which has zero derivative at the origin), so the one-variable function $y \mapsto f_x(0,y)=0$ is continuous.

But $f$ isn't even continuous at the origin, hence not differentiable there.