I must be doing something wrong here.
Let $f: \mathbb R^2 \to \mathbb R$ be defined as
$$f(x,y)= \begin{cases} 3x+4y, &\text{if } xy \neq 0 \\ 0, &\text{if } xy =0 \end{cases}$$
i.e. $f$ is $0$ on the $x-$axis and on the $y-$axis whereas $f(x,y)=3x+4y$ everywhere else.
I want the check the differentiability at $(0,0)$.
$\bullet$ Clearly, $\ f$ is continuous.
$\bullet$ Also, I think that $f_x=0=f_y$, the partial derivatives are just the function restricted to $x-$axis and $y-$axis. I found the picture below on the web, it gives a geometric interpretation of partial derivatives.
$\bullet$ So, the partial derivatives exist and continuous. Then, $f$ is differentiable at $(0,0)$.
$\bullet$ However, I think that $f$ is not differentiable at the origin, because... Umm... The directional derivative is not continuous. A little direction change may yield a huge change in directional derivative.
Is $f$ differentiable at the origin? What is wrong about my reasoning?
$$$$ $$$$
Edit: So, isn't $f_x$ defined as the slope of the tangent line to the curve that is given by the intersection of the graph of $f$ and the $xz-$plane? (that is the red line in the above picture)
