Continuity and differentiability here involve two dimensional limits that fail to exist at (0,0), which can be verified (I think) using the cubic path. The directional derivatives are one dimensional limits along straight lines (or rays) starting at (0,0), so they only involve paths (t,mt) and (0,t).
Added in edit:
As you say, the partial derivatives at $(0,0)$ are both $0$ since $f$ is identically $0$ along both $x$ and $y$ axes, and so $grad(f)(0,0)=0$. Hence, if $f$ were differentiable at $(0,0)$, all directional derivatives at $(0,0)$ would equal $0$ by the usual formula (the dot product for the directional derivative at a point of differentiability).
To compute the directional derivative at $(0,0)$ in an arbitrary direction (i.e. unit vector) $(c,s)$ with $c <> 0$, find the following limit as $t -> 0$ :
$lim (f(0+tc,0+ts)-f(0,0))/t = lim ( (ct)(st)^2 )/( (t^2)(c^2+(t^4)(s^6))t ) = s^2/c$
which is obviously not $0$ at arbitrary unit vectors $(c,s)$. But the limit, and hence the directional derivative, DOES exist for all directions (the directions with $c=0$ are along the y axis and those directional derivatives are just (+/-) the partial w.r.t $y$ at the origin-- in this case, both are 0). [By the way $c$ and $s$ stand for cosine and sine].
Finally, if we approach the origin along the path $(t^3,t)$ then the one dimensional limit of the function itself is
$lim (t^5)/(t^6 + t^6) = lim 1/(2t)$ which is NOT $0$ and so $f$ is not continuous at the origin (since the full two dimensional limit at $(0,0)$ can't be $0$ if the one dimensional path limits don't all come out to $0$)
Sorry, I don't know how to format better. Also, I have forgotten the definition of differentiability -- but a consequence of it is the chain rule, from which is derived the formula for directional derivative in terms of the partial derivatives (and I think is what you were referring to with "linear..."). Since the formula fails here, the function must not be differentiable at the origin (and it fails to be continuous at the origin as well, which also implies non-differentiability).
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 Answer
Theorem 25.1 in Rockafellar's Convex Analysis states that if the subdifferential of $f$ at $x$ is a singleton, then $f$ is differentiable at $x$.
Let $u,v\in \partial f(x)$. Note that for any direction $y$, if $\alpha >0$ $$\langle y,u \rangle = \frac 1{\alpha} \langle y,\alpha u \rangle \leq \frac{f(x+\alpha y)-f(x)}{\alpha}$$ and if $\alpha <0$ $$\langle y,u \rangle = \frac 1{\alpha} \langle y,\alpha u \rangle \geq \frac{f(x+\alpha y)-f(x)}{\alpha}$$ Letting $\alpha \to 0$ yields $\langle y,u \rangle = f'(x,y)$.
Replacing $u$ with $v$, $\langle y,v \rangle = f'(x,y) = \langle y,u \rangle$. Hence $\forall y\in \mathbb R^n, \langle y,u-v\rangle =0$ and $u-v=0$. $\partial f(x)$ is thus a singleton.