The tangent line to the function's graph at any point \((a,b)=\left(a,a+\dfrac2a\right)\) has slope
\(f'(x)=1-\dfrac2{x^2}\)
and thus has the equation
\(y-a-\dfrac2a=\left(1-\dfrac2{a^2}\right)(x-a)\implies y=\left(1-\dfrac2{a^2}\right)x+\dfrac4a\)
Any such line that passes through (3,3) will then satisfy
\(3=\left(1-\dfrac2{a^2}\right)3+\dfrac4a\mplies\dfrac6{a^2}=\dfrac4a\)
\(a\) clearly can't be 0, so we can divide both sides by \(\dfrac1a\) to end up with
\(\dfrac6a=4\implies a=\dfrac64=\dfrac32\)
So the only point that answers the question is \(\left(\dfrac32,\dfrac32+\dfrac2{\frac32}\right)=\left(\dfrac32,\dfrac{17}6\right)\)
