Prove f(x) can take all real values

You could restrict yourself to the interval (-3,3) for x.
Also note the if
y = 2(3x+1)/3(x^2-9)
can take on any value in the interval, so can
y = 3(x^2-9)/2(3x+1)

Not quite true. Knowing that f(x) takes every value on (-inf, inf) doesn't guarantee that 1/f(x) ever takes the value 0.

Also, I don't think there's actually any advantage to flipping the RHS here: you want to show there are solutions for any choice of y, so the original equation becomes 3y(x^2-9) = 2(3x+1) which is still just a quadratic in x.

I don't know what "techniques described" the OP is talking about; I think the sketch you suggest is more "informative", but in an exam where I wanted to be sure of the marks, I think I'd go this algebra route.

Yeah just remembered from a-level, what you're meant to do is turn it into a quadratic in x, find the descrim and show that it is >=0 for all values of y .

Is this correct?


$3y(x^2-9) = 2(3x+1)$ rearranges to $3yx^2-6x-(27y+2)=0$

Finding the discriminant

$324y^2+24y+36\geq0$

Taking out a factor of 12

$27y^2+2y+3\geq0$

Completing the square

$27(y^2+1/27y)^2+(80/27)>0$

since

$27(y^2+1/27y)^2\geq0$ for all real y, and $80/27>0$