The following theorem is Theorem 6.2 on p. 50 in "Analysis on Manifolds" by James R. Munkres.
Theorem 6.2. Let $A$ be open in $\mathbb{R}^m$. Suppose that the partial derivatives $D_jf_i(x)$ of the component functions of $f$ exist at each point $x$ of $A$ and are continuous on $A$. Then $f$ is differentiable at each point of $A$.
The following exercise is Exercise 3 on p. 54 in "Analysis on Manifolds" by James R. Munkres.
Exercise 3. Show that the proof of Theorem 6.2 goes through if we assume merely that the partials $D_j f$ exist in a neighborhood of $a$ and are continuous at $a$.
I noticed that the following proposition holds.
My Proposition A.
Let $A\subset\mathbb{R}^m$; let $f:A\to\mathbb{R}^n$. Suppose $A$ contains a neighborhood of $a$.
Let $j_1,\dots,j_n\in\{1,\dots,m\}$.
Suppose that for each $i\in\{1,\dots,n\}$ and $j\in\{1,\dots,m\}\setminus\{j_i\}$, $D_jf_i(x)$ exists at each point $x$ of a neighborhood of $a$ and is continuous at $a$.
Suppose that $D_{j_i}f_i(a)$ exists for each $i\in\{1,\dots,n\}$.
Then $f$ is differentiable at $a$.
To prove my proposition A, it is sufficient to prove my proposition B since the following Theorem 5.4 on p.46 in "Analysis on Manifolds" by James R. Munkres holds.
Theorem 5.4 Let $A\subset\mathbb{R}^m$; let $f:A\to\mathbb{R}^n$. Suppose $A$ contains a neighborhood of $a$.
$\cdots$
(a) The function $f$ is differentiable at $a$ if and only if each component function $f_i$ is differentiable at $a$.
My Proposition B.
Let $A\subset\mathbb{R}^m$; let $f:A\to\mathbb{R}$. Suppose $A$ contains a neighborhood of $a$.
Let $j_1\in\{1,\dots,m\}$.
Suppose that for each $j\in\{1,\dots,m\}\setminus\{j_1\}$, $D_jf(x)$ exists at each point $x$ of a neighborhood of $a$ and is continuous at $a$.
Suppose that $D_{j_1}f(a)$ exists.
Then $f$ is differentiable at $a$.
I prove my proposition B for the case in which $m=2$ and $j_1=2$.
Proof
Let $a:=(b,c)$.
$$f(b+h, c+k)-f(b,c)=f(b+h,c+k)-f(b,c+k)+f(b,c+k)-f(b,c)\\=\frac{\partial{f}}{\partial{x}}(b+th,c+k)h+\left(\frac{\partial{f}}{\partial{y}}(b,c)k+o(k)\right)$$ for some $t\in (0,1)$.
$$f(b+h, c+k)-f(b,c)-\left(\frac{\partial{f}}{\partial{x}}(b,c)h+\frac{\partial{f}}{\partial{y}}(b,c)k\right)\\=\left(\frac{\partial{f}}{\partial{x}}(b+th,c+k)-\frac{\partial{f}}{\partial{x}}(b,c)\right)h+o(k)$$ for some $t\in (0,1)$.
So,
$$\left(\frac{\partial{f}}{\partial{x}}(b+th,c+k)-\frac{\partial{f}}{\partial{x}}(b,c)\right)\frac{h}{\sqrt{h^2+k^2}}+\frac{o(k)}{k}\cdot\frac{k}{\sqrt{h^2+k^2}}\to 0$$ as $\sqrt{h^2+k^2}\to 0$.
Does my proposition A really hold?
