Question related to first order partial derivatives If The funtion $f: \Bbb R^2 \to \Bbb R$ has directional derivatives in all directions at each point in $\Bbb R^2$ then the function $f$ has first order partail derivatives at each point in $\Bbb R^2$ 
How can I prove this? Please do not downvote because of not saying any idea or my tried. I cannot produce any idea to prove this. Please help me thank you 

Hints: Use the definitions.
What does it mean for $f$ to have directional derivatives in all direction at each point in $\Bbb R^2$?
It means that for all vectors $(k,l)\in \Bbb R^2$ and for all points $(x,y)\in \Bbb R^2$ the following limit exists: 
$$\lim \limits_{h\to 0}\left(\frac{f((x,y)+h(k,l))-f(x,y)}{h}\right) \tag 1$$
Now set $(k,l)=(1,0)$ and $(k,l)=(0,1)$ in $(1)$ and  check the definition of partial derivative.
Partial answer (pun): 
$\begin{align} \displaystyle \lim \limits_{h\to 0}\left(\frac{f((x,y)+h(1,0))-f(x,y)}{h}\right)&=\lim \limits_{h\to 0}\left(\frac{f(x+h,y)-f(x,y)}{h}\right)=\frac{\partial f}{\partial x}(x,y) \end{align}$
The last equality only holds because the first limit exists and because it is equal to the second limit in the chain of equalities above.
