Let $f(x)=\sqrt x-\dfrac{x\log x}{x-1}$ $(\log$ is the natural logarithm$)$.
Prove there exists an extension $F(x)$ of $f(x)$ in $x_0=1$, where at least $F(x)\in C^2((0,+\infty))$
My attempt $$\lim_{x\to1}f(x)=\lim_{y\to0}f(1+y)=\lim_{y\to0}1-\frac{(y+1)\log(1+y)}{y}=\lim_{y\to0}1-(y+1)=0$$ Then $f$ can be extended: $$F(x)=\begin{cases}f(x)&x\neq1\\0&x=1\end{cases}$$ I check the differentiability of $F$ in $1$: $$\lim_{x\to1} \frac{F(x)-F(1)}{x-1}=\lim_{y\to0} \frac{F(1+y)}{y}=0\implies F'(1)=0$$
I know that $F(x)\in C^1.$ Then I try to check if $F(x)\in C^2.$
$$\lim_{x\to1} \frac{F'(x)-F'(1)}{x-1}=\lim_{x\to1} \frac{f'(x)}{x-1}=\frac1{12}\implies F''(1)=\frac1{12}$$
Is this a right approach?
Yes, your approach if correct. As an inspiration, try to expand the function formally around $x=1$: $$ f(x)=\frac{1}{24}(x-1)^2-\frac{1}{48}(x-1)^3+o((x-1)^3). $$ Hence $f(0)=0$ and $f'(0)=0$ are necessary conditions, and you can check directly that the first one is also sufficient.
You could also try to use the so-called Converse to Taylor's Theorem, see here.