Specifically, I only have to show this for $f:R^2\rightarrow R$. We know that $f_{,1}$ and $f_{,2}$ exist and are bounded in a region about (0,0), and we need to show that f is continuous at (0,0).
I have been struggling with continuity of functions in more than one variable. I have very little intuition, and I can't seem to get a start on this.
Pick a ball $B=B(0, r)$ on which the partials are bounded by say $M$, and $x=(x_1,x_2)\in B$. The mean value theorem for real functions gives, for some $c$ between $0\in \mathbb{R}$ and $x_2$, $$ |f(0,0)-f(0,x_2)|= |\partial_2 f(0,c)||x_2| \leq M|x_2|\leq M|x|. $$ Similarly we get, for some $d$ between $0\in \mathbb{R}$ and $x_1$, $$ |f(0,x_2)-f(x_1,x_2)|=|\partial_1 f(d,x_2)||x_1| \leq M|x|. $$ Combining this $$ |f(0,0)-f(x)|\leq |f(0,0)-f(0,x_2)|+|f(0,x_2)-f(x_1,x_2)| \leq 2M|x|. $$ Since $M$ doesn't depend on $x$ the result follows.