Cross-multiplication is applicable when have two fractions or rational expressions equated to each other.  It is method wherein we multiply the denominator towards the numerator on the other side. This will helps to simplifying the equation by getting rid of the fraction form on each side.

For a given equation:  `a/b=c/d` where b and are nonzero, we may cross-multiply to simplify it as:  `a*d = c*b` .

Applying cross-multiplication on the given equation `9/(x^2-6x+9)=(3x)/(x^2-3x)` , we get:

`9*(x^2-3x)=(3x)*(x^2-6x+9)`

Apply distribution property.

`9x^2-27x=3x^3-18x^2+27x`

Subtract `9x^2` from both sides of the equation.

`9x^2-27x-9x^2=3x^3-18x^2+27x-9x^2`

`-27x=3x^3-27x^2+27x`

Add `27x`  on both sides of the equation.

`-27x+27x=3x^3-27x^2+27x+27x`

`0=3x^3-27x^2+54x`

or `3x^3-27x^2+54x=0`

Factor out the greatest common factor (GCF) `3x` .

`(3x)(x^2-9x+18)=0`

Apply `x^2-9x+18= (x-3)(x-6)` , we get:

`3x(x-3)(x-6)=0`

Apply zero-factor property to solve for x by equating each factor to `0` .

`3x=0`

`(3x)/3 =0`

`x=0`

`x-3=0`

`x-3+3=0+3`

`x=3`

`x-6=0`

`x-6+6=0+6`

`x=6`

Possible values of `x=0,3,6` .

To check for extraneous solution, plug-in each x on `9/(x^2-6x+9)=(3x)/(x^2-3x)` .

Note: Any value divided by `0` results to undefined value.

An undefined result implies the x value is an extraneous solution.

Let `x=0`  on `9/(x^2-6x+9)=(3x)/(x^2-3x)` .

`9/(0^2-6*0+9)=?(3*0)/(0^2-3*0)`

`9/(0-0+9)=?0/(0-0)`

`9/9=?0/0`

1=? undefined   FALSE

Let `x=3`  on `9/(x^2-6x+9)=(3x)/(x^2-3x)` .

`9/(3^2-6*3+9)=?(3*3)/(3^2-3*3)`

`9/(9-18+9)=?9/(9-9)`

`9/0=?9/0`

undefined=? undefined   FALSE

Let `x=6`  on `9/(6^2-6x+9)=(3x)/(x^2-3x)` .

`9/(6^2-6*6+9)=?(3*6)/(6^2-3*6)`

`9/(36-36+9)=?9/(36-18)`

`9/9=?18/18`

`1=1 `             TRUE

Therefore, the `x=0` and `x=3` are the extraneous solutions.

The `x=6` is the real exact solution of the given equation `9/(x^2-6x+9)=(3x)/(x^2-3x)` .