 Solution

The $i$th partial derivative is ${\partial }_{i}f(x_0)={\lim }_{t\to 0}[f(x_0+t{e}_i)-f(x_0)]/t$. 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 $L:{\mathbb{R}}^n\to \mathbb{R}$ withFor $v=(v_1,v_2)$ with $v_1\ne 0$, 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 every directional derivative exists and is zero.

Along the curved path $(x,y)=(t^3,t)$, however, $f(t^3,t)=1$ for $t\ne 0$, while $f(0,0)=0$. Hence the function is not continuous at the origin. Differentiability implies continuity, so it is not differentiable there either. Direction-by-direction limits alone do not control general approaches to a point.