I've been self studying real analysis lately and were trying to solve the above problem from my textbook. The actual problem mentions a maximum or minimum, but I figured proving the statement for a maximum would suffice as the other case is essentially the same but with different sign. I think I managed to prove the claim, but I was wondering whether someone here could double check it. My proof goes as follows:
Assume, on the contrary, that some partial derivative at $x=a$ is not zero, say, $\partial f/ \partial x_j (a) = m$. Without loss of generality, assume that $m>0$. Let $f_j: \mathbb{R} \rightarrow \mathbb{R}$ be the $j$-th component function of $f$. Then $f_j$ is differentiable and $f_j '(a)=m>0$, meaning that there exists some $L>0$ such that $f_j$ is increasing on $[a-L, a+L]$. On the other hand, by the definition of the $j$-th partial derivative at $x=a$, we can write $$f(a+L)=f(a)+L \cdot m + r(L),$$ where $\frac{|r(L)|}{L} \rightarrow 0$ as $L \rightarrow 0$. In particular, we may take $L$ small enough such that $|r(L)|<L \cdot m$. Using this on the previous equation then gives us $$f(a+L)=f(a)+L \cdot m + r(L) \geq f(a)+L \cdot m -|r(L)|>f(a)+L \cdot m-L \cdot m=f(a).$$ We thus proved that $f$ does not attain a maximum at $x=a$.
This is essentially totally right (well done!), the idea being that one can simply look at the one-variable functions where only one coordinate is changing at a time.
The imperfection is that $f'_j(a)>0$ doesn't actually imply that there's an entire interval around $a$ on which $f_j$ is increasing (think of $f_j$ involving some extremely wiggly function such as $x_j^2\sin\frac1{x_j}$). But it's easy to modify your argument to get around this gap:
Since $\lim_{\ell\to0} r(\ell)/\ell = 0$, choose $\delta>0$ such that $|r(\ell)|/\ell < m$ for $|\ell| < \delta$. Then in particular, when $\ell$ is positive, $f(a+\ell) = f(a) + \ell m + r(\ell) > f(a)$, which is the desired contradiction.