= Solution
The $i$th <partial derivative> is $\partial_if(x_0)=\lim_{t\to0}[f(x_0+te_i)-f(x_0)]/t$. The <directional derivative> in direction $v$ is the same limit with $v$ in place of $e_i$, using the specified vector rather than normalizing it. <Differentiability> means that there is a linear map $L:\mathbb R^n\to\mathbb R$ with
$$
f(x_0+h)=f(x_0)+Lh+o(\|h\|).
$$
For $v=(v_1,v_2)$ with $v_1\ne0$, the difference quotient at the origin is $t v_2^3/v_1$, tending to zero. If $v_1=0$, the <function> is identically zero on the directional line, again giving derivative zero. Thus \b[every <directional derivative> exists and is zero].
Along the curved path $(x,y)=(t^3,t)$, however, $f(t^3,t)=1$ for $t\ne0$, while $f(0,0)=0$. Hence the <function> is \b[not <continuous>] at the origin. <Differentiability> implies <continuity>, so it is \b[not differentiable] there either. Direction-by-direction limits alone do not control general approaches to a point.
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