Here's a real-analytic example: Let $g(t) =1/(1+t^2).$ Define
$$f(x,y) = \begin{cases} xg((y-x^2)/(x^2+y^2)^2), (x,y) \ne (0,0) \\
0, (x,y) =(0,0).\end{cases}$$
Then $f$ is real-analytic on $\mathbb {R}^2 \setminus \{(0,0)\},$ $f$ is continuous at $(0,0),$ and $D_uf(0,0) = 0$ for all unit vectors $u.$ Because $f(x,x^2) = xg(0) = x\ne o(x,x^2),$ $f$ is not differentiable at $(0,0).$
Best to have some examples where it is possible to calculate things. In this one, we switch to polar coordinates to see that it is Gateaux differentiable, indeed all directional derivatives are zero. However, it is not Frechet differentiable, as a path that is not a straight line gives a poorly behaved quotient.
Take $$ g(0,0) = 0, \; \mbox{otherwise} \; \; g(x,y) = \frac{x^5 \; y^5}{x^{12} + y^8} $$
Added, August 20, 2017: it occurred to me that I had not proved $g$ continuous. Lagrange multipliers says that the maximum of $|x^5 y^5|,$ for fixed $x^{12} + y^8,$ occurs when $2 y^8 = 3 x^{12}.$ Fiddle with it a little, we get
$$ |g| \leq C \; \; \sqrt[\color{red}{24}] {x^{12} + y^8}, $$ where
$$ C = \frac{ 2^{\left( \frac{5}{12}\right)} 3^{\left( \frac{5}{8}\right)}}{ 5^{\left( \frac{25}{24}\right)}}.$$
So $g$ is continuous at the origin.
Next, we write the function itself in polar coordinates,
$$ g = r^2 \left( \frac{\cos^5 \theta \; \; \sin^5 \theta}{r^4 \cos^{12} \theta + \sin^8 \theta} \right) $$
Along either coordinate axis, we get constant zero. Otherwise, with the function defined as zero at the origin, for Gateaux we simply take $g/r,$ giving
$$ \frac{g}{r} \; = \; \; r \; \; \left( \frac{\cos^5 \theta \; \; \sin^5 \theta}{r^4 \cos^{12} \theta + \sin^8 \theta} \right) $$
For any fixed $\theta$ direction (not on the x,y axes), this gives limit $0$ as $r \rightarrow 0,$ therefore directional derivative zero.
Everything looks promising in polar coordinates, giving Gateaux.
But then, along path $x = t^2, y = t^3,$ we find the quotient
$$ \frac{g(t^2, t^3)}{\sqrt{t^4 + t^6}} \geq \frac{1}{4|t|} $$
when $|t| \leq \sqrt 3,$ so $g$ is not Frechet.
Best Answer
Let $f \colon \mathbb R^2 \to \mathbb R$ be given by $$f(x,y) = \begin{cases} x & x \ne 0, y = x^2, \\ 0 & \text{else}.\end{cases}$$ This function is directionally differentiable (with a linear derivative) and continuous in $(0,0)$.
With your definition of Gâteaux differentiability, you can even use any norm on $\mathbb R^2$, e.g., $$f(x,y) = \sqrt{x^2 + y^2}$$ or $$f(x,y) = |x| + |y|.$$