Proof: If the $f_x(x,y)$ exists on $(x_0,y_0)$, $f_y(x,y)$ is continuous on the $(x_0,y_0)$, then $f(x,y)$ is differentiable 
Proof: If the $f_x(x,y)$ exists on $(x_0,y_0)$, $f_y(x,y)$ is continuous on the $(x_0,y_0)$, then $f(x,y)$ is differentiable on $(x_0,y_0)$.

I can not do it because the condition is too weak. Thank you in advance.
 A: Hint: One needs to show that $f$ has a derivative $f'(x_0, y_0) : \mathbb{R}^2 \to \mathbb{R}$, which should have the form
$$a\mathbf{e}_1 + b\mathbf{e}_2 \mapsto af_x(x_0, y_0) + bf_y(x_0, y_0).$$
So, we need to show that the limit of the following vanishes:
$$\lim_{h \to 0} \frac{\Big(f(x_0 + h_1, y_0 + h_2) - f(x_0,y_0)\Big) - \Big(h_1f_x(x_0, y_0) + h_2f_y(x_0, y_0)\Big)}{\lVert h \rVert} = 0. $$
Now we should reason component-wise—but first let us assume $\lVert h \rVert < \varepsilon$, where $\varepsilon > 0$ is such that $f_y$ is exists on $B_{\varepsilon}(x_0, y_0)$ and is continuous at $(x_0, y_0)$. Now we may write
$$f(x_0 + h_1, y_0 + h_2) - f(x_0,y_0) = \Big(f(x_0 + h_1, y_0 + h_2) -  f(x_0 + h_1, y_0)\Big) + \Big(f(x_0 + h_1, y_0) - f(x_0,y_0)\Big).$$
By the Mean Value Theorem, and existence of $f_y$, there exists some $c \in (0, h_2)$ such that
$$f(x_0 + h_1, y_0 + h_2) -  f(x_0 + h_1, y_0) = h_2f_y(x_0 + h_1, y_0 + c).$$
Now we may use the continuity of $f_y$ to compare this to $h_2f_y(x_0, y_0)$. Note that we don't need to know that $f_x$ is continuous, or even exists anywhere else, to compare the term $f(x_0 + h_1, y_0) - f(x_0,y_0)$ to $h_1f_x(x_0, y_0)$.
After such comparisons it should be clear the limit is $0$.
