# Continuity of an extended rate of variation?

Let $f:\mathbb R\to \mathbb R$ be a $C^1$ function.

Then Prove or disprove that $g:\mathbb R^2\to \mathbb R$ define by $$g(x,y) = \begin{cases}\frac{f(x)-f(y)}{x-y}&\text{if x\neq y}\\ f'(x)&\text{if x= y}\end{cases}$$ is a continuous function.

Note that $C^1$ Continuity must be crucial since a blatant counter example given by the function $$f(x) = \begin{cases}x^2\sin \frac{1}{x}&\text{if x\neq 0}\\ 0&\text{if x= 0}\end{cases}$$ which differentiable on $\mathbb R$ but not $C^1$ because the derivative is not continuous at $x=0.$

first we see that g is continuous at $(x,y)$ such that $x\neq y$ now let $x_{0} = y_{0}$ and let $(x,y)$ converge to $(x_{0},y_{0})$ we have $\frac{f(y)-f(x)}{y-x}= {f}'(c)$ such that $x\leq {c} \leq y$ . and the conclusion follow since $f$ is $C^{1}$ continuous