No function can have a graph with positive measure or even positive inner measure, since every function graph has uncountably many disjoint vertical translations, which cover the plane.
Meanwhile, using the axiom of choice, there is a function whose graph has positive outer measure. The construction is easiest to see if one assumes that the Continuum Hypothesis is true, so let me assume that.
To begin, note first that there are only continuum many open sets in
the plane, since every such set is determined by a
countable union of basic open balls with rational center
and rational radius. Next, it follows that the number of
$G_\delta$ sets is also continuum, since any such set is
determined by a countable sequence of open sets, and
$(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}$.
Thus, we may enumerate the $G_\delta$ sets in the plane as
$A_\alpha$ for $\alpha\lt \aleph_1$ (using CH). Build a
function $f:\mathbb{R}\to\mathbb{R}$ by transfinite
induction. At any stage $\alpha\lt \aleph_1$, we have
the approximation $f_\alpha$ to $f$, and we assume that it
has been defined on only $\alpha$ many points. Given
$f_\alpha$, consider the $G_\delta$ set $A_\alpha$. If we
can extend $f_\alpha$ to a function $f_{\alpha+1}$ by
defining it on one more point $x$, so that
$(x,f_{\alpha+1}(x))$ is outside $A_\alpha$, then do so.
Otherwise, $A_\alpha$ contains the complement of
countably many vertical lines in the plane, and thus has
full measure.
After this construction, extend the resulting function if
necessary to a total function $f:\mathbb{R}\to\mathbb{R}$.
It now follows that the graph of $f$ is not contained in
any $G_\delta$ set with less than full measure. Thus, the
graph has full outer measure.
Now, finally, the same construction works without CH, once you realize that any $G_\delta$ set containing the complement of fewer than continuum many vertical lines has full measure.
Basically, just do what the hint says. The idea is to cover the curve (i.e. the graph of your function $f:\mathbb R \to\mathbb R$) by rectangles such that the sum of their areas (measures) is smaller than $\epsilon>0$ and do so for every $\epsilon>0$. This would show that $m(C)<\epsilon$ for every $\epsilon>0$ and thus $m(C)=0$.
Let's do this.
Lemma. Let $g:[a,b]\to\mathbb R$ be a continuous function and let $\Gamma_g = \{(x,g(x))|\;x\in[a,b]\}$ be the graph of $g$. Then $m(\Gamma_g)=0$.
Proof. The function $g$ is uniformly continuous, since $[a,b]$ is compact. Let $\epsilon>0$. Then there exists a $\delta>0$, such that for each pair of points $x,y\in [a,b]$ such that $|x-y|<\delta$, the inequality $|g(x)-g(y)|<\epsilon$ holds. We may assume without loss of generality that $\delta<b-a$ (since if some $\delta\geq b-a$ works for this choice of $\epsilon$, any other number $0<\delta_1<b-a$ would also work.)
Let $n\in\mathbb N$ be the smallest natural number such that $n\delta> b-a$. Then there exist numbers $a=x_0<x_1<\ldots<x_n=b$ on the interval $[a,b]$, such that for each $i=1,2,\ldots,n$ we have $|x_i-x_{i-1}|<\delta$. (Proving this is an easy exercise.) For each $i=1,2,\ldots,n$ let $y_i=g(x_i)$.
Now, for each $i=1,2,\ldots,n$ and each point $z\in[x_{i-1},x_i]$, we have $g(z)\in[y_i-\epsilon,y_i+\epsilon]$ because for such $z$ the inequality $|z-x_i|<\delta$ holds, and thus by definition of $\delta$, we also have the inequality $|g(z)-y_i|<\epsilon$. This implies that $\Gamma_g\subseteq\bigcup_{i=1}^n[x_{i-1},x_i]\times[y_i-\epsilon,y_i+\epsilon]$ and therefore $$m(\Gamma_g)\leq\sum_{i=1}^n m([x_{i-1},x_i]\times[y_i-\epsilon,y_i+\epsilon])\leq\sum_{i=1}^n 2\delta\epsilon = 2\delta\epsilon n$$ Since $n$ is the smallest natural number for which $n\delta>b-a$, we must also have $b-a+\delta\geq n\delta$ (since otherwise we would have $b-a+\delta<n\delta$ which would imply $b-a<(n-1)\delta$, which would contradict our choice of $n$). This means that $2\delta\epsilon n\leq 2\epsilon(b-a+\delta)<4\epsilon(b-a)$ since $\delta<b-a$.
So, we showed that for every $\epsilon>0$ the inequality $m(\Gamma_g)<4\epsilon (b-a)$ holds. This is only possible if $m(\Gamma_g)=0$ and thus the proof is complete. $\square$
Now, to show this for $f:\mathbb R\to\mathbb R$ notice that the graph of $f$ (defined by $\Gamma_f=\{(x,f(x))|\;x\in\mathbb R\}$) is the union of graphs of restrictions to $[-n,n]$. More precisely: define for each $n\in\mathbb N$ a function $f_n:[-n,n]\to\mathbb R$ by the formula $f_n(x) = f(x)$. Then $$\Gamma_f=\bigcup_{n=1}^\infty\Gamma_{f_n}.$$ Therefore, $$m(\Gamma_f)\leq\sum_{n=1}^\infty m(\Gamma_{f_n})=\sum_{n=1}^\infty 0 = 0$$ and thus $m(\Gamma_f)=0$. We are done.
Best Answer
Let $f : \Bbb{R}^n \to \Bbb{R}$ and define
$$ G := \{(x,f(x)) \mid x \in \Bbb{R}^n\}, $$
the graph of $f$. Observe that $G$ is closed, hence measurable.
By the Fubini-Tonelli theorem, applied to the characteristic function/indicator function $\chi_G$ of $G$, we see
$$ \lambda_{n+1}(G) = \int_{\Bbb{R}^{n+1}} \chi_G (x,y) d(x,y) = \int_{\Bbb{R}^n} \int_\Bbb{R} \chi_G (x,y) \, dy \, dx = \int_{\Bbb{R}^n} 0 \, dx = 0, $$
where the step before the last used that $\chi_G (x,y) \neq 0$ implies $(x,y) \in G$ and hence $y = f(x)$, so that $\chi_G (x,y) = 0$ for all $y \neq f(x)$ and hence in particular for almost all $y \in \Bbb{R}$, so that $\int_\Bbb{R} \chi_G (x,y) dy = 0$.
EDIT: Note that we did use the continuity of $f$ only to conclude that $G$ is indeed measurable. This is the case as soon as $f$ is (Borel/Lebesgue) measurable.