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