Continuity of one partial derivative implies differentiability Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ be a function such that the partial derivatives with respect to $x$ and $y$ exist and one of them is continuous. Prove that $f$ is differentiable.

In short: the problem reduces to the easy case when $f$ depends solely on one variable. See the greyish box below for the formula that does the reduction.
It suffices to show that $f$ is differentiable at $(0,0)$ with the additional assumption that $\frac{\partial f}{\partial x}(0,0)=\frac{\partial f}{\partial y}(0,0)=0$. First pass from $(x_0,y_0)$ to $(0,0)$ by considering the function $g(x,y)=f(x+x_0,y+y_0)$. Then work on $h(x,y)=g(x,y)-x\frac{\partial f}{\partial x}(0,0)-y\frac{\partial f}{\partial y}(0,0)$.
So let us assume assume that $\frac{\partial f}{\partial x}$ exists and is continuous on $\mathbb{R}^2$ (only continuity in an open neighborhood of $(0,0)$ is really needed for the local argument), that $\frac{\partial f}{\partial y}$ exists at $(0,0)$, and that $\frac{\partial f}{\partial x}(0,0)=\frac{\partial f}{\partial y}(0,0)=0$. We need to show that $f$ is differentiable at $(0,0)$. Note that the derivative must be $0$ given our assumptions.
Now observe that for every $x,y$, we have, by the fundamental theorem of calculus:

$$
f(x,y)=f(0,y)+\int_0^x \frac{\partial f}{\partial x}(s,y)ds.
$$

I let you check properly that $(x,y)\longmapsto f(0,y)$ is differentiable at $(0,0)$ with zero derivative, using only $\frac{\partial f}{\partial y}(0,0)=0$. For the other term, just note that it is $0$ at $(0,0)$ and that for every $0<\sqrt{x^2+y^2}\leq r$
$$
\frac{1}{\sqrt{x^2+y^2}}\Big|\int_0^x \frac{\partial f}{\partial x}(s,y)ds\Big|\leq \frac{|x|}{\sqrt{x^2+y^2}}\sup_{0\leq \sqrt{x^2+y^2}\leq r}\Big| \frac{\partial f}{\partial x}(s,t)\Big|\leq \sup_{0\leq \sqrt{x^2+y^2}\leq r}\Big| \frac{\partial f}{\partial x}(s,t)\Big|.
$$
By continuity of $\frac{\partial f}{\partial x}$ at $(0,0)$, the rhs tends to $0$ when $(x,y)$ tends to $(0,0)$. This proves that the function $(x,y)\longmapsto \int_0^x \frac{\partial f}{\partial x}(s,y)ds$ is differentiable at $(0,0)$ with zero derivative. And this concludes the proof.
