The function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ has, at the point $(0,0)$, for $v \in \mathbb{R}^2\setminus \{(0,0)\}$, the directional derivative given by $$ \frac{\partial f}{\partial v} (0,0) = \frac{|v_1| v_2}{|v_1| + |v_2|} + 3v_1 - 2v_2. $$ Compute the partial derivatives at $(0,0)$ and determine whether $f$ is differentiable at $(0,0)$.
So I think I got the partial derivatives (by substituting with unit vectors): $$ \frac{\partial f}{\partial e_1} (0,0) = \frac{|1| \cdot 0}{|1| + 0} + 3 \cdot 1 - 2 \cdot 0 = 3, $$ $$ \frac{\partial f}{\partial e_2} (0,0) = \frac{|0| \cdot 1}{|0| + 1} + 3 \cdot 0 - 2 \cdot 1 = -2. $$ But I'm not so sure how one can show whether $f$ is differentiable in $(0,0)$ or not... my intuition tells me it's not differentiable.. but I feel like I don't have enough info. I was maybe thinking of showing that $f$ doesn't have all directional derivatives at $(0,0)$ but I couldn't figure it out.
If $f$ is differentiable in $a$ then you know that for $v = (v_1,v_2)$ you should have
$$ \frac{\partial f}{\partial v} (a)= (\nabla f (a)) \cdot v = v_1 \frac{\partial f}{\partial e_1}(a) + v_2 \frac{\partial f}{\partial e_2}(a)$$
In particular this means that if the function $f$ is differentiable in $a$ then the function $$ v \mapsto \frac{\partial f}{\partial v} (a)$$ must be a linear function, which is not the case for our function $f$ at $a = (0,0)$.
As Frank suggested, considering the vector $v = (1,1)$ shows the function is not differentiable. Indeed if the function is differentiable then you should have
$$ \frac{\partial f}{\partial v}(0,0) = 1\frac{\partial f}{\partial e_1}(0,0) + 1 \frac{\partial f}{\partial e_2}(0,0) = 3-2 = 1.$$
But this is not the case :
$$ \frac{\partial f }{\partial v} (0,0) = \frac{1}{1+1} + 3 - 2 = \frac{1}{2} + 1 = \frac{3}{2}.$$