You need to remember that the image of the function represents the range of the function, hence, replacing each value of x from domain of function, in equation of function, yields the image of the function.

The problem provides the information that the image of the function belongs to interval `(0,6)` , such that:

`0 < (x^2-ax+4)/(x^2+4) < 6 => {((x^2-ax+4)/(x^2+4) > 0),((x^2-ax+4)/(x^2+4) < 6):}`

Solving the first inequality, yields:

`(x^2-ax+4)/(x^2+4) > 0`

Since the denominator is always positive, yields:

`x^2-ax+4 > 0`

The quadratic expression is strictly positive if the parabola does not touch the x axis, hence, the quadratic equation has no real roots.

You need to consider the condition for a quadratic does not have any real roots, such that:

`Delta < 0 => b^2 - 4ac < 0`

Identifying a,b,c yields:

`a=1,b=-a,c=4`

`(-a)^2 - 4*1*4 < 0 => a^2 - 4^2 < 0`

You need to attach the equation, such that:

`a^2 - 4^2 = 0 => (a - 4)(a + 4) = 0`

Using zero product rule, yields:

`a - 4 = 0 => a = 4`

`a + 4 = 0 => a = -4`

The quadratic expression `a^2 - 4^2 < 0` for `a in (-4,4)` .

Solving the bottom inequality, yields:

`(x^2-ax+4)/(x^2+4) < 6 => (x^2-ax+4)/(x^2+4) - 6 < 0`

`(x^2 - ax + 4 - 6x^2 - 24)/(x^2+4)< 0`

`(-5x^2 - ax - 20)/(x^2+4)< 0 => (5x^2 + ax + 20)/(x^2+4)> 0`

Since the denominator is always positive, yields:

`5x^2 + ax + 20 > 0 => Delta < 0`

`Delta = a^2 - 20^2 < 0 => (a - 20)(a + 20) < 0 => a in (-20,20)`

You need to evaluate the solution to simultaneous inequalities, such that:

`a in (-4,4) nn (-20,20) => a in (-4,4)`

**Hence, evaluating a, under the given condition, `Im f = y in (0,6)` , yields **`a in (-4,4).`