The slope of the tangent to 9x - 4x*ln y = 3 at (1/3 ,1) has to be determined.

Use implicit differentiation to find y'.

`9 - 4*ln y - 4x*(1/y)*y' = 0`

=> ` 4x*(1/y)*y' = 9 - 4*ln y`

=> `y' = (y*(9 - 4*ln y))/(4*x)`

At the point (1/3, 1) the slope of the tangent is the value of y' at that point. Here y' = `(1*(9 - 4*ln 1))/(4*(1/3))`

=> (9*3)/4

=> 27/4

The equation of the tangent is `(y - 1)/(x - 1/3) = 27/4`

=> `4y - 4 = (27x)/4 - (27/4)*(1/3)`

=> `4y - 4 = (27x)/4 - 9/4`

=> `16y - 16 = 27x - 9`

=> `27x - 16y + 7 = 0 `

**The equation of the tangent to `9x - 4x*lny = 3` at `(1/3 ,1) ` is **`27x - 16y + 7 = 0 `