When #x !=0# and #x!=2#, we can write

#f(x)=(x(x-2))/(x(x-2)(x-2))=(cancel(x)(cancel(x-2)))/(cancel(x)(cancel(x-2))(x-2))=1/(x-2)#.

Therefore, the graph of #f# is exactly the same as the graph of #g(x)=1/(x-2)# when #x!=0# and #x!=2#. The graph of #g# only has a discontinuity at #x=2#, meaning the discontinuity of #f# at #x=2# is non-removable (there's a vertical asymptote there).

On the other hand, #g(0)=-1/2#, meaning that #lim_{x->0}f(x)=-1/2#, meaning the discontinuity of #f# at #x=0# can be "removed" by re-defining #f# in a piecewise way there to be #f(0)=-1/2#. In other words, the discontinuity of #f# at #x=0# is removable.

The graph shown below is the graph of #g#, which is essentially the graph of the original (non-piecewise) #f#, except that the point #(x,y)=(0,-1/2)# would be "missing" from the graph of the original #f#.

graph{1/(x-2) [-10, 10, -5, 5]}