It is not hard to show that if the limit given in the post exists, then the derivative exists, and is equal to the given limit. Just let $k=h$, and rewrite the expression as
For the converse, suppose $f'(x)$ exists. Rewrite our expression as
Choose an $\epsilon \gt 0$. Because $f'(x)$ exists, there is a $\delta$ such that if $|t|\lt \delta$, then
It follows that if $h$ and $k$ are $\lt \delta$, then the differential quotients in $(1)$ are each within $\epsilon$ of $f'(x)$. Thus $(1)$ is equal to
where $\epsilon_1$ and $\epsilon_2$ are non-negative quantities $\lt \epsilon$.
But $h/(h+k)$ and $k/(h+k)$ are both less than $1$. (This little fact is key: we could run into trouble using two points on the same side of $x$.)
It follows that if $h$ and $k$ are less than $\delta$, then $(1)$ differs from $f'(x)$ by less than $\epsilon$. So the limit as $h$, $k$ independently go to $0^+$ exists.