Let $f:\mathbb{R} \rightarrow \mathbb{R} $ of class $C^{1}$, show that the function $F:\mathbb{R^{2}} \rightarrow \mathbb{R} $ is differentiable if $x \neq y$. Also show that that if $f$ is of class $C^{2}$, then F is differentiable in all $\mathbb{R^{2}}$
The function F is given by:
$ F=\begin{cases} \frac{f(y)-f(x)}{y-x}, x\neq y \\ f'(x), x=y \end{cases} $
You can tell that for $x\neq y$ the function $F$ is differentiable because it is a quotient of two differentiable functions and the denominator is not zero.
To prove that if $f\in C^2$ then $F$ is differentiable, it is enough to prove that $F$ is differentiable on the line $x=y$. The other points we already know.
Let's look at the point $(x_0,x_0)$ on that line. At that point, for $h_1\neq h_2$,
$$Q=\frac{\left|F(x_0+h_1,x_0+h_2)-F(x_0,x_0)-D\right|}{|h_1|+|h_2|}=\frac{\left|\frac{f(x_0+h_2)-f(x_0+h_1)}{h_2-h_1}-f'(x_0)-D\right|}{|h_1|+|h_2|}$$
Since $f\in C^2$ we can write $f(x_0+h)=f(x_0)+f'(x_0)h+f''(x_0)h^2/2+o(|h|^2)$ the last fraction for $Q$ is equal to
$$\frac{\left|f''(x_0)(h_1+h_2)-D+\frac{o(|h_2|^2)-o(|h_1|^2)}{h_2-h_1}\right|}{|h_1|+|h_2|}$$
So, taking $D$ to be the function $D(h_1,h_2)=f''(x_0)(h_1+h_2)/2$, which is linear in $(h_1,h_2)$ we get that the quotient $Q$ above tends to $0$ as $(h_1,h_2)\to0$ along $h_1\neq h_2$.
Finally, along $h_1=h_2$ we have that the quotient $Q$ is $$\frac{\left|f'(x_0+h_1)-f'(x_2)-D\right|}{|h_1|+|h_2|}$$
By taking $D$ as before, which on $h_1=h_2$ is becomes $f''(x_0)h_1$, we get that $Q$ tends to $0$ as $h_1\to 0$.
Combining the two limits we get that tending $(h_1,h_2)$ in any fashion, with the choice of differential $D(h_1,h_2)=f''(x_0)(h_1+h_2)/2$ the function $F$ satisfies the definition of differentiable at $(x_0,x_0)$.