Do functions with the same gradient differ by a constant?

[Spoiler warning, this is more than a hint. I wanted to show this method because it avoids working with components.]

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First suppose that $h:mathbb R^n to mathbb R$ is differentiable and that $nabla h(x) = 0$ for all $x in mathbb R^n$. I'll prove that $h$ is constant. Suppose (for a contradiction) that there exist points $a$ and $b$ in $mathbb R^n$ such that $h(a) neq h(b)$. Let $z:[0,1] to mathbb R$ be the function defined by $$ z(t) = h(a + t(b - a)). $$ Note that $z$ is continuous on $[0,1]$ and differentiable on $(0,1)$ and that $z(0) neq z(1)$. By the mean value theorem, there exists a number $c$ such that $0 < c < 1$ and $$ z'(c) = z(1) - z(0) neq 0. $$ But, by the chain rule, $$ z'(c) = langle nabla h(a + c(b -a)), b - a rangle $$ which is $0$ because we are assuming that $nabla h(x) = 0$ for all $x$ in $mathbb R^n$. This is a contradiction. Therefore $h$ is constant.

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Next, to solve the original problem, let $h = f - g$ and apply the above result.

Yes that's correct. Solve each ($1$-dimensional) equation $partial{f}/partial{x_i} = partial{g}/partial{x_i}$

HINT: Integrate both sides of

$df=frac{partial f}{partial x}dx+frac{partial f}{partial y}dy+cdots= vec{nabla f}cdot vec{dl}=vec{nabla g}cdot vec{dl}=dg$

EDIT: As pointed out by @peek-a-boo, this only works if the functions in question have an integrable derivative (so for example a sufficient condition is for $f$ to be $C^1$; i.e continuously differentiable).

If $nabla f=nabla g$, then $frac{partial f}{partial x_k}=frac{partial g}{partial x_k}$ for all $kin{1,ldots,n}$. Thus there exists $c_k(x_1,ldots,x_{k-1},x_{k+1},ldots,x_n)$ such that $f=g+c_k$. And, for $lneq k$, we have $$ frac{partial c_k}{partial x_l}=0 $$ and thus $dc_k=0$ and $c_k$ is a constant. The value of $c_k$ then does not depend of $k$ since for all $k,l$, $f-g=c_k=c_l$. Thus there exists a constant $c$ such that $f=g+c$.