A function $f:\mathbb{R}^2 \to \mathbb{R}$ is differentiable at a point $p$ if there is a linear map $L:\mathbb{R}^2 \to \mathbb{R}$ such that $$\lim_{h \to 0}\frac{\left|f(p+h)-f(p)-L(h)\right|}{|h|} = 0$$ If the function is differentiable, then $L$ is the function $L(x,y) = \dfrac{\partial f }{\partial x}\big|_p x +\dfrac{\partial f}{\partial y}\big|_p y$.
If all you know is that the partial derivatives exist, you do not even know that the function is continuous, let alone is well approximated by a tangent plane. For instance $f(x,y) = 0$ if $x=0$ or $y=0$ and $f(x,y) = 1$ otherwise. The partial derivatives of $f$ at $(0,0)$ are all $0$, but the tangent plane is a really crappy approximation to $f$ off of the coordinate axes.
The example you were looking at is a little harder to visualize, but think about what happens to $f$ on lines other than horizontal and vertical ones.
In other words, morally, for a function to be differentiable at $(a,b)$ we need that $f(a+\Delta x,b+\Delta y) \approx f(a,b) + \frac{\partial f}{\partial x} \Delta x + \frac{\partial f}{\partial y} \Delta y$. This talks about approximating $f$ in all directions around $(a,b)$, whereas existence of the partial derivatives only means you have good approximations in two directions (the coordinate directions).
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).
Best Answer
Existence of every directional derivative of $f$ at a point $p$ does not even imply $f$ is continuous at $p$. (!) For example, let $E$ denote the parabola $y = x^{2}$ with the origin removed, and let $f$ be the characteristic function of $E$, namely $$ f(x, y) = \begin{cases} 1 & y = x^{2},\ x \neq 0, \\ 0 & \text{otherwise.} \end{cases} $$
Understanding this example thoroughly should clarify your doubts about your assigned question. The point is, for every line $\ell$ through the origin, $\ell$ contains an open interval about $(0, 0)$ that misses $E$.[*] Consequently, $f$ is locally constant along every line through the origin, so its directional derivative in an arbitrary direction vanishes.
Conceptually, controlling the behavior of a function along lines through a point need not give control over non-lines through that point. That's why limits at $p$ are defined by looking at the behavior of $f$ in open disks about $p$.
[*] Precisely, for every line $\ell$ through $(0, 0)$, there exists an $r > 0$ such that $\ell$ does not meet $E$ inside the open disk of radius $r$ centered at the origin.