Continuous function with linear directional derivatives=>Total differentiability?

The answer is "no".

Let $f:\mathbb R^2\to\mathbb R$ be defined by $f(x,x^2)=x$ for all $x\in \mathbb R$ and $f(x,y)=0$ if $y\neq x^2$. This function $f$ is continuous at $0$ with $f(0)=0$. For any fixed direction $v$, we have $f(tv)=0$ if $t$ is small enough; so $\partial_v f(0)$ exists with $\partial_vf(0)=0$. But $f$ is not differentiable at $0$ because the only possible (total) differential is $L=0$ and we don't have $f(x,y)=o(\Vert (x,y)\Vert)$ as $(x,y)\to 0$.

The answer is "no."

Let $f: R^2 \to R$ be defined by $f(x,y)= \frac{x^3y}{x^4+y^2}$ for $(x,y) \neq (0,0)$ and $f(0,0) = 0$. This function is continuous; its directional derivative is defined at each point in every direction; and at each point its directional derivative is a linear function of the direction. But, you can check that $f$ is not differentiable at $(0,0)$ by approaching $(0,0)$ along $y=x^2$.

See Foundations of Modern Analysis by Dieudonne, Vol. 1 Chapter VIII Section 4.