Is it possible that a directional derivative exists on a discontinuous point? So I have this function:

$$
f(x,y) =
\begin{cases}
\dfrac{xy^2}{x^2 + y^2},  & \text{if $(x,y) \neq (0,0)$} \\
0, & \text{if $(x,y) = (0,0)$}
\end{cases}
$$

And I am asked to calculate the directional derivative at $(0,0)$, along a vector $(a,b)$. I used the limit definition and got to a result, but then realized that maybe it should not exist since the function is not continuous. Is this true?
 A: The function $f$ in your post is continuous at $(0,0)$ since $|f(x,y)|\leqslant|x|$ for every $(x,y)$ and $|x|\to0$ when $(x,y)\to(0,0)$. The directional derivative of $f$ at $(0,0)$ in the direction $(\cos\theta,\sin\theta)$ is $$\lim_{r\to0}\frac1r\frac{r^3\cos\theta\sin^2\theta}{r^2}=\cos\theta\sin^2\theta$$

For an example of a function not continuous at $(0,0)$ such that every directional derivative at $(0,0)$ exists, consider $g(x,y)=0$ for every $(x,y)$ in $\mathbb R^2$ with the exception that $g(x,x^2)=1$ for every $x\ne0$.
Then $g(x,0)=0$ for every $x$ hence the directional derivatives of $g$ at $(0,0)$ in the direction of $(\pm1,0)$ exist and are both $0$. For every direction $(u,v)$ with $v\ne0$, $g(tu,tv)=0$ for every $|t|$ small enough (say, every $|t|\leqslant |v|/(u^2+1)$) hence the directional derivative of $g$ at $(0,0)$ in the direction of $(u,v)$ exists and it is $0$.
But naturally, this function $g$ is not continuous at $(0,0)$ since $g(0,0)=0$ and $\lim\limits_{x\to0}g(x,x^2)=1\ne0$.
A: Since
$$
|f(x,y)| \leq |x| \frac{y^2}{x^2+y^2} \leq |x|,
$$
the function is continuous at the origin. Anyway, all you can deduce from the existence of a directional derivative at some point is that the restriction of the function to the straight line passing through that point in the given direction is continuous. The function may be discontinuous, though.
A: If the directional derivative of a function exists at some point, then it is not necessarily true that the function is continuous at that point.
