TITLE: Continuous function with linear directional derivatives=>Total differentiability?
QUESTION [13 upvotes]: As in the title: 
If $f\colon\mathbb{R}^n\to\mathbb{R}$ is continuous in $x$ and has directional derivatives $\partial_vf(x)=L(v)\,\forall v\in\mathbb{R}^n$, where $L$ is linear, does this imply that $f$ is totally differentiable?

REPLY [7 votes]: 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.