This question has the following hint:
Find a linear function $L(x,y)$ such that $f(x+h,y+k)-f(x,y)-L(h,k)=hk$. Note that $|hk|\leq l^2$ where $l:=max\{|h|,|k|\}$. Does this help you establish that a certain limit is zero? To establish that the limit is zero, provide an $\epsilon-\delta$ proof.
My initial post questioned the linear function part; hence the first couple comments. I'm now adding my proposed solution that I'd like to have critiqued.
Let $h,k\in\mathbb{R}$ such that $|hk|\leq l^2$ where $l:=\mbox{max}\{|h|,|k|\}$, and suppose $L(h,k)=kx+hy$. Thus, $$\frac{|f(x+h,y+k)-f(x,y)-L(h,k)}{||(h,k)||}=\frac{|(x+h)(y+k)-xy-(kx+hy)|}{\sqrt{h^2+k^2}}=\frac{|hk|}{\sqrt{h^2+k^2}}.$$ Then, note that $$|h|^2\leq h^2+k^2\Rightarrow|h|\leq\sqrt{h^2+k^2}\Rightarrow\frac{|h|}{\sqrt{h^2+k^2}}\leq1\Rightarrow\frac{|hk|}{\sqrt{h^2+k^2}}\leq|k|.$$ Hence, $$\frac{|f(x+h,y+k)-f(x,y)-L(h,k)}{||(h,k)||}=\frac{|hk|}{\sqrt{h^2+k^2}}\leq|k|.$$ So, $$\lim_{(h,k)\rightarrow(0,0)}\frac{|f(x+h,y+k)-f(x,y)-L(h,k)}{||(h,k)||}=0.$$ Therefore, by the definition of derivative, $f$ is differentiable everywhere.
It seems to me that an $\epsilon-\delta$ proof is unnecessary, but maybe I'm implicitly making an assumption that needs proved?