Your function is continuous in $B^{n}=\{x\in\mathbb{R}^{n}: \|x\|\leq 1\}$
Hence $f$ must attain a maximum and a minimum in $B^{n}$. All we have to show is that there exist a maximum or a minimum in $int (B^{n})$, where $int$ is interior.
Indeed, $f$ cannot assume a minimum in a point $u\in S^{n-1}$ because by hypothesis, the function $f$ restricted to the set $\{tu:\ t\in\mathbb{R}\}$ is increasing in a neighbourhood of $u$. Then the minimum is attained in a point $u\in int(B{^{n}})$.
We can define a function $\phi: \mathbb{R} \to \mathbb{R}$ where $ \phi (t)=f(a+tv)$, whith $f(a)$ a "local" minimum point. As there is all directional derivates $\phi$ is derivable and $\phi(0)$ is minimum point. Then we have $\phi'(0)=0$ and $\phi(0)=f'(a;v)$.