Singularity of sin x / x Is the definition of singularity here is incomplete?
http://mathworld.wolfram.com/SingularPoint.html
Take $f(x, y) = y - {\sin x\over x} = 0$.
$f_y = 1$ for all points so the condition $f_x = f_y = 0$ can never be true.
But
http://mathworld.wolfram.com/RemovableSingularity.html
says $f$ is singular when $x=0$.
 A: You can define a plane curve by the implicit equation
$$F(x,y)=0$$ where
$$F(x,y)=\begin{cases}y-1,x=0\\y-\frac{\sin(x)}{x},x\neq 0\end{cases}$$
for $(x,y)\in\mathbb{R}^2$. Note that $F$ is smooth, this is because as You observed
$$F_y=1$$ for all points and because the function
$$\hat{f}(x)=\begin{cases}1,x=0\\ \frac{\sin(x)}{x}\end{cases}$$
is smooth. To see that the first derivative exists use the rule of De L'Hospital twice:
$$\lim_{h\rightarrow 0,h\neq 0}\frac{\hat{f}(0)-\hat{f} (h)}{h}=\lim_{h\rightarrow 0,h\neq 0}\frac{1-\frac{\sin(h)}{h}}{h}=\lim_{h\rightarrow 0,h\neq 0}\frac{h-\sin(h)}{h^2}\underbrace{=}_{De L'H}$$
$$\lim_{h\rightarrow 0,h\neq 0}\frac{1-\cos(h)}{2h}\underbrace{=}_{De L'H}\lim_{h\rightarrow 0,h\neq 0}\frac{\sin(h)}{2}=0.$$
So $$\hat{f}'(x)=\begin{cases}0,x=0\\\frac{\cos(x)x-\sin(x)}{x^2},x\neq 0\end{cases}.$$
Note that $\lim_{x\rightarrow 0}\frac{\cos(x)x-\sin(x)}{x^2}=0$. One would have to prove the existence of derivatives of any order to show smoothness. The latter essentially means that $x=0$ is a $\textbf{removable singularity}$ of the function $f(x)=\frac{\sin(x)}{x}$. Now 
$$F_x(x_0,y_0)=\begin{cases}0,x=0\\-\frac{\cos(x)x-\sin(x)}{x^2},x\neq 0\end{cases}$$
implies $F_x(0,y_0)=0$ for every $y_0\in\mathbb{R}$ but since $F_y(0,y_0)=1$ for all $y_0\in\mathbb{R}$ none of these points ( in particular not $(0,1)$, which is the only one actually on the considered curve) is a singular point to the plane curve given by $F(x,y)=0$. 
