If $y=f(x)=\frac{x+2}{x-1}$, then it is incorrect to say:
$\textbf{(A)\ }x=\frac{y+2}{y-1}\qquad\textbf{(B)\ }f(0)=-2\qquad\textbf{(C)\ }f(1)=0\qquad$
$\textbf{(D)\ }f(-2)=0\qquad\textbf{(E)\ }f(y)=x$
$f(1)=\frac{3}{0}$, which is undefined. Hence, the (in)correct answer is $\boxed{\textbf{(C)}\ f(1)=0}$.
We can verify that the other statements are valid.
$\textbf{(A).\ } y=\frac{x+2}{x-1}\implies xy-y=x+2\implies x(y-1)=y+2\implies x=\frac{y+2}{y-1}$
$\textbf{(B).\ } f(0)=\frac{(0)+2}{(0)-1}=-2$
$\textbf{(D).\ } f(-2)=\frac{(-2)+2}{(-2)-1}=\frac{0}{-3}=0$
$\textbf{(E).\ } f(y)=\frac{\frac{x+2}{x-1}+2}{\frac{x+2}{x-1}-1}=\frac{3x}{3}=x$