How you would approach this problem depends on whether you
are taking college algebra or calculus.

If you are taking algebra and have not studied calculus,
then

The equation of a line through (0,6) which is the y-intercept
and b=6, is

y = mx+b or

y = mx+6

We solve that system of equations

y = mx+6
y = 1/x

mx+6 = 1/x

Multiply through by x

  mx²+6x = 1
mx²+6x-1 = 0

A line is tangent to a curve when there is 
just one point of intersection.  That is,
when the quadratic equation has discriminant = 0

discriminant = B²-4AC = 6²-4(m)(-1) = 0

6²-4(m)(-1) = 0
      36+4m = 0
         4m = -36
          m = -9

So  y = mx+b becomes:
    y = -9x+6

which shows that that is a tangent line.

--------------------------------

Regardless of which you are taking, you will have to find the point of intersection:

Solve the system:

           y = 1/x
           y = -9x+6

Set the right sides equal:

         1/x = -9x+6

Multiply through by x

           1 = -9x²+6x
    9x²-6x+1 = 0
(3x-1)(3x-1) = 0
        3x-1 = 0
          3x = 1
           x = 

Substitute in

    y = -9x+6
    y = -9(+6
    y = -3+6
    y = 3

So the point of intersection, which is the point of tangency, is:

      (,3)

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If you are studying calculus, find the derivative of

y = 1/x
Rewrite as
y = x-1
 = -1x-2
 = 

Substitute  for x in  

 =  =  = -9

which is the same as the slope of the line.

Therefore the line is tangent to the curve at their
point of intersection. 

Edwin