Suppose that $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ exist in an open ball $B((a,b),R), (R>0)$, and are continuous in $(a,b)$. Then $f$ is differentiable in $(a,b)$.
Proof:
We want to show that $$ \lim_{(x,y)\to (a,b)} \frac{f(x,y)-f(a,b)-\frac{\partial f}{\partial x}(a,b)(x-a)-\frac{\partial f}{\partial y}(a,b)(y-b)}{\sqrt{(x-a)^2+(y-b)^2}}=0.$$
The argument can be rewritten as
$$ \frac{f(x,y)-f(a,y)-\frac{\partial f}{\partial x}(a,b)(x-a)}{\sqrt{(x-a)^2+(y-b)^2}}+\frac{f(a,y)-f(a,b)-\frac{\partial f}{\partial y}(a,b)(y-b)}{\sqrt{(x-a)^2+(y-b)^2}}.$$ We will show that $$ \lim_{(x,y)\to (a,b)} \frac{f(x,y)-f(a,y)-\frac{\partial f}{\partial x}(a,b)(x-a)}{\sqrt{(x-a)^2+(y-b)^2}}=0.$$ (Similarly for the second term).
By applying the mean value theorem* on the function $f(\cdot ,y)$, $y$ fixed, we get $$ f(x,y) - f(a,y) = (x-a) \frac{\partial f}{\partial x}(\xi,y), \quad \xi\in ]a,x[.$$ $\dots$
*My question: Why can we apply the mean value theorem here? The conditions should be 1) $f(\cdot,y)$ is differentiable over $]a,x[$, and 2) continous over $[a,x]$.
Condition (1) is fulfilled, since $B((a,b),R)$ (with $b$ fixed) becomes an open interval on which $\frac{\partial f}{\partial x}$ (= $f'$) exists.
I can't find a reasonable explanation for the second condition to be fulfilled. Any ideas?