If $f$ is differentiable in $B(a)$ and $f(x) \leq f(a)$ for all $x$ in $B(a)$, then $\nabla f(a) = 0$ 
Assume $f$ is differentiable at each point of an n-ball $B(a)$. Prove that if $f(x) \leq f(a)$ for all $x$ in $B(a)$, then $\nabla {f(a)} = 0.$

I had my proof, but I'm not sure it is correct.
Proof:
Since f is differentiable at each point of the n-ball B(a), meaning
$$\lim_{h \to 0} \frac{f(a+hy)-f(a)}{h} = \nabla f(a) \cdot y$$
, where y is an arbitrary unit vector.
From the mean value theorem, we know that 
$$\lim_{h \to 0} \frac{f(a+hy)-f(a-hy)}{h} = \nabla f(c) \cdot y$$
for some c where $||c|| < r$.
Since
$$\lim_{h \to 0} \frac{f(a+hy)-f(a)}{h} = \nabla f(c) \cdot y = - \lim_{h \to 0} \frac{f(a)-f(a-hy)}{h}$$
Since the RHS of the the first equation is 0, we have $\nabla f(a) = 0$.
So, is there any mistake of any suggestion about the point that I can improve mathematically or about the way that I wrote ?

Let $y \in R^n$ an arbitrary vector and let be $\alpha: (-\epsilon,\epsilon) \to B(a)$ given $\alpha(h) = a + hy$. Then the differentiable function $g: (-\epsilon,\epsilon) \to R$  where $g = f \circ \alpha$ has critical value in $h=0$ and therefore $$ 0 = g´(0) = \nabla f . y$$, how $y$ is arbitrary, this implies that $\nabla  f = 0$
