Official solution
Given \[y = {{x + 6} \over {\left( {x - 2} \right)\left( {x - 2} \right)}}\]
At y-axis, x = 0 \[ \Rightarrow \] y = 1
On differentiating, we get
\[{{dy} \over {dx}} = {{\left( {{x^2} - 5x + 6} \right)\left( 1 \right) - \left( {x + 6} \right)\left( {2x - 5} \right)} \over {{{\left( {{x^2} - 5x + 6} \right)}^2}}}\]
\[{{dy} \over {dx}} = 1\] at point (0, 1)
\[ \therefore \] Slope of normal = -- 1
Now equation of normal is y -- 1 = --1 (x -- 0)
\[ \Rightarrow \] y -- 1 = -- x
x + y = 1 ......(1)
By checking each option you can see point \[\left( {{1 \over 2},{1 \over 2}} \right)\] satisfy equation (1).