 Solution

The $i$th partial derivative is the limit ${\partial }_{i}f(x_0)={\lim }_{t\to 0}[f(x_0+t{e}_i)-f(x_0)]/t$. For a nonzero vector $v$, use the directional convention ${\partial }_{v}f(x_0)={\lim }_{t\to 0}[f(x_0+tv)-f(x_0)]/t$, without normalizing $v$. Differentiability means $f(x_0+h)=f(x_0)+Lh+o(\Vert h\Vert )$ for a linear map $L=df(x_0)$.

For the given example take $v=(v_1,v_2)$. If $v_1\ne 0$, then $f(tv)=t^2{v}_2^3/v_1$, so its directional difference quotient tends to zero. If $v_1=0$, the function vanishes along that line. Thus $\boxed{{\partial }_{v}f(0)=0\text{ for every }v\ne 0}$. However along $(x,y)=(t^3,t)$ with $t\ne 0$, one has $f(x,y)=1$, while $f(0,0)=0$. Hence 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.