= Solution
The $i$th <partial derivative> is the limit $\partial_if(x_0)=\lim_{t\to0}[f(x_0+te_i)-f(x_0)]/t$. For a nonzero vector $v$, use the directional convention $\partial_vf(x_0)=\lim_{t\to0}[f(x_0+tv)-f(x_0)]/t$, without normalizing $v$. <Differentiability> means $f(x_0+h)=f(x_0)+Lh+o(\|h\|)$ for a <linear map> $L=df(x_0)$.
For the given example take $v=(v_1,v_2)$. If $v_1\ne0$, then $f(tv)=t^2v_2^3/v_1$, so its directional difference quotient tends to zero. If $v_1=0$, the function vanishes along that line. Thus $\boxed{\partial_vf(0)=0\text{ for every }v\ne0}$. However along $(x,y)=(t^3,t)$ with $t\ne0$, one has $f(x,y)=1$, while $f(0,0)=0$. Hence \b[the function is neither <continuous> nor <differentiable> at zero]. Existence of all <directional derivatives>, even with a linear dependence on the direction, does not by itself imply total <differentiability>.
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