First, switch y and x in your equation:
x = 3 log_2(4y) - 2
Now, solve this equation for y:
x = 3 log_2 (4y) - 2
<=> x + 2 = 3 log_2(4y)
<=> (x+2)/3 = log_2(4y)
The inverse function of log_2(a) is 2^a, so apply this operation to both sides of the equation to get rid of the logarithm:
<=> 2^((x+2)/3) = 2^(log_2(4y))
<=> 2^((x+2)/3) = 4y
Let's simplify the expression on the left side using the power rules a^n * a^m = a^(n+m) and a^(n*m) = (a^n)^m:
2^((x+2)/3) = 2^(x/3 + 2/3) = 2^(x/3) * 2^(2/3) = 2^(x/3) * (2^2)^(1/3) = 4^(1/3) * 2^(x/3)
Let's get back to our equation:
2^((x+2)/3) = 4y
<=> 4^(1/3) * 2^(x/3) = 4y
<=> 4^(1/3)/4 * 2^(x/3) = y
<=> 4^(-2/3) * 2^(x/3) = y
You are done. The only thing left to do is replacing y with f^(-1)(x) for a more formal notation:
for
f(x) = 3 log_2(4x) - 2,
the inverse function is
f^(-1)(x) = 4^(-2/3) * 2^(x/3).
Hope that this helped!