We need
#(u/v)'=(u'v-uv')/(v^2)#
Let's calculate the first derivative
#f(x)=x^2/(x-2)^2#
#u=x^2#, #=>#, #u'=2x#
#v=(x-2)^2#, #=>#, #v'=2(x-2)#
#f'(x)=(2x*(x-2)^2-x^2*(2(x-2)))/(x-2)^4#
#=((x-2)(2x(x-2)-2x^2))/(x-2)^4#
#=(2x^2-4x-2x^2)/(x-2)^3#
#=(-4x)/(x-2)^3#
#f'(x)=0#, when #x=0#
We build our first chart
#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaaaaa)##0##color(white)(aaaaaaaaaa)##2##color(white)(aaaaaa)##+oo#
#color(white)(aaaa)##-4x##color(white)(aaaaaa)##+##color(white)(aaaa)##0##color(white)(aaaa)##-##color(white)(aaaa)##||##color(white)(aaa)##-#
#color(white)(aaaa)##(x-2)^3##color(white)(aaa)##-##color(white)(aaaa)##0##color(white)(aaaa)##-##color(white)(aaaa)##||##color(white)(aaa)##+#
#color(white)(aaaa)##f'(x)##color(white)(aaaaa)##-##color(white)(aaaa)##0##color(white)(aaaa)##+##color(white)(aaaa)##||##color(white)(aaa)##-#
#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##↘##color(white)(aaa)##0##color(white)(aaaa)##↗##color(white)(aaa)##||##color(white)(aaa)##↘#
The critical point is when #x=0#
The intervals of increasing are #x in (0,2)# and decreasing are #x in (-oo,0) uu (2,+oo)#
Now, we calculate the second derivative
#u=-4x#, #=>#, #u'=-4#
#v=(x-2)^3#, #=>#, #v'=3(x-2)^2#
So,
#f''(x)=(-4(x-2)^3+4x*3(x-2)^2)/(x-2)^6#
#=(-4(x-2)^2((x-2)-3x))/(x-2)^6#
#=-4(x-2-3x)/(x-2)^4#
#=-4(-2x-2)/(x-2)^4#
#=(8(x+1))/(x-2)^4#
Therefore,
#f''(x) =0# when #x=-1#, this is the inflexion point
The chart is as follows
#color(white)(aaaa)##I##color(white)(aaaaaa)##(-oo,-1)##color(white)(aaaa)##(-1,2)##color(white)(aaaa)##(2,+oo)#
#color(white)(aaaa)##f''(x)##color(white)(aaaaaaa)##-##color(white)(aaaaaaaaa)##+##color(white)(aaaaaaaa)##+#
#color(white)(aaaa)##f''(x)##color(white)(aaaaaaa)##nn##color(white)(aaaaaaaaa)##uu##color(white)(aaaaaaaa)##uu#
The curve is concave is up when # x in (-1,2) uu (2,+oo)#
The curve is concave down when # x in (-oo,-1)#
Also,
#f''(0)>0#, this is a minimum
graph{x^2/(x-2)^2 [-11.36, 11.14, -1.58, 9.67]}