How would I prove that if $f: \mathbb R^2 \to \mathbb R$ is a function such that
$$\lim_{(x,y)\to(a,b)} f(x,y) = L$$
and for every $y_0 \in \mathbb R$
$$ \lim_{x\to a} f(x,y_0) = L'_{y_0}$$
and for every $x_0 \in \mathbb R$
$$ \lim_{y\to b} f(x_0,y) = L''_{x_0}$$
then
$$ \lim_{x\to a}\left(\lim_{y\to b} f(x,y)\right) = \lim_{y\to b}\left(\lim_{x\to a} f(x,y)\right) = L$$
You cannot prove that the two-variable limit equals the iterated limits even if they both exist, since the two-variable limit may fail to exist even if both iterated limits exists and are equal. For example, take $f(x,y) = \frac{xy}{x^2+y^2}$, with $a=b=0$. The iterated limits both exist: $$\lim_{x\to 0}\left(\lim_{y\to 0}\frac{xy}{x^2+y^2}\right) = \lim_{x\to 0}\frac{0}{x^2} = 0,$$ $$\lim_{y\to 0}\left(\lim_{x\to 0}\frac{xy}{x^2+y^2}\right) = \lim_{y\to 0}\frac{0}{y^2} = 0.$$ But the limit does not exist as $(x,y)\to(0,0)$. If you approach $(0,0)$ along the line $y=x$ you get the limit as $x\to 0$ of $\frac{x^2}{2x^2}$, which goes to $\frac{1}{2}$, not $0$. So the two-variable limit does not exist, even though the two iterated limits both exist and are equal.
If you assume that the two-sided limit exists and is equal to $L$, and then you want to prove that the iterated limits will therefore both exist and be equal, then you might have a better shot. Start by thinking about the definitions.
Added later: And then look at Robin Chapman's example to see that it does not quite work out. So you don't get implications either way.
This is a rejoinder to Arturo's comment that if the two-variable limit exists then both iterated limits exist. Define $$f(x,y)=\left\{\begin{array}{cl} (x^2+y^2)\sin(1/x+1/y)&\textrm{if $xy\ne0$,}\\ 0&\textrm{if $xy=0$.} \end{array}\right.$$ Then certainly $f$ is continuous at $(0,0)$; that is $$\lim_{(x,y)\to(0,0)}f(x,y)=0.$$ But for any nonzero $x$, $$\lim_{y\to0}f(x,y)$$ doesn't exist (the function is wildly oscillatory for $y$ near zero). So $$\lim_{x\to0}\left(\lim_{y\to0}f(x,y)\right)$$ is meaningless.
