Suppose $f$ satisfies $f_x(x,y) = f_y(x,y) = 0$. Then $f$ is constant. 
Suppose $f$ is a function with $f_x(x,y)=f_y(x,y)=0$ for all $(x,y)$. Then show that $f(x,y)=c$, a constant.

I had the following method in mind. Fix $h,k$ to be any particular point in the $X-Y$ plane. Then, since partial derivatives exist, by the second FTC,
$$f(h,k)-f(h,0)=\int_0^k f_y(h,y) dy = 0 \implies f(h,k) = f(h,0)$$
and $$f(h,0) - f(0,0) = \int_0^h f_x(x,0) dx = 0 \implies f(h,0) = f(0,0)$$
$$\implies f(x,y)=f(0,0)=c$$
What is wrong in this proof? I haven't assumed $f$ is differentiable either (though it is since the partial derivatives are continuous)
 A: There is nothing wrong with your proof.
Indeed, the conditions that $f_x = f_y = 0$ for all $\vec{x}$ means that all partial derivatives exist and are continuous. Therefore $f$ is differentiable. And further, the gradient $\nabla f$ is identically $0$.
So this is a direct analog of the 1-variable statement that if $g: \mathbb{R}\longrightarrow \mathbb{R}$ is differentiable and $g'(t) = 0$ for all $t$, then $g$ is constant. [If you recall, this is a direct corollary of the mean value theorem].
You might be interested in a different (but equally valid as your own) proof relying more closely on the associated $1$-variable result.
We can define $g(t) = f(\vec{a} + t(\vec{b} - \vec{a}))$ (where $f$ is our function above), then $g$ is a differentiable function (since $f$ is differentiable). Then $f(\vec{b}) - f(\vec{a}) =g(1) - g(0) = g'(\xi)$ for some $\xi \in (0,1)$ by the mean value theorem. As $g' \equiv 0$, we see that $f(\vec{b}) = f(\vec{a})$. This can be done for any two points $\vec{a}$ and $\vec{b}$, so $f$ is constant.
