There are a couple ways to go about this.
By definition of inverse function, if [tex]j^{-1}[/tex] exists, then
[tex]j\left(j^{-1}(x)\right) = x[/tex]
All this means is that if you provide some input [tex]x[/tex], plug it into [tex]j^{-1}(x)[/tex], then plug that into [tex]j(x)[/tex], you end up recovering the original input [tex]x[/tex] that you started with.
We can also do this in the opposite direction, so
[tex]j^{-1}(j(x)) = x[/tex]
So now if [tex]j(x)[/tex] is invertible, then
[tex]j(x) = -\dfrac23 x - 9 \implies j\left(j^{-1}(x)\right) = -\dfrac23 j^{-1}(x) - 9 = x[/tex]
Solve for [tex]j^{-1}[/tex].
[tex]-\dfrac23 j^{-1}(x) - 9 = x[/tex]
[tex]\dfrac23 j^{-1}(x) = -x - 9[/tex]
[tex]j^{-1}(x) = -\dfrac32 (x+9)[/tex]
[tex]\boxed{j^{-1}(x) = -\dfrac32 x - \dfrac{27}2}[/tex]
Another way to do it is by algebraically substituting [tex]x[/tex] within the definition of [tex]j(x)[/tex]
Starting with the given definition
[tex]j(x) = -\dfrac23 x - 9[/tex]
we replace [tex]x[/tex] with some new expression in a new variable [tex]y[/tex] that makes the right side reduce to just [tex]y[/tex]. To demonstrate, if we multiply [tex]-\frac23 x[/tex] by [tex]-\frac32[/tex], we end up with a coefficient of 1. So let [tex]x=-\frac32 y[/tex]. Then
[tex]j\left(-\dfrac32 y\right) = -\dfrac23 \left(-\dfrac32 y\right) - 9 = y - 9[/tex]
As you can see, this doesn't completely reduce to [tex]y[/tex], so we're not done. To adjust for this, we just add 9 to [tex]y[/tex], so our ultimate substitution is [tex]x=-\frac32(y+9)[/tex]. Then
[tex]j\left(-\dfrac32 (y+9)\right) = -\dfrac23 \left(-\dfrac32 (y+9)\right) - 9 = (y+9) - 9 = y[/tex]
Now applying [tex]j^{-1}[/tex] to both sides,
[tex]j^{-1}\left(j\left(-\dfrac32 (y+9)\right)\right) = j^{-1}(y)[/tex]
[tex]\implies \boxed{j^{-1}(y) = -\dfrac32 y - \dfrac{27}2}[/tex]
(we can freely swap out [tex]y[/tex] for [tex]x[/tex] at this point)
Yet another way, since the function here is quite simple, we can just think about the action of [tex]j[/tex] on a given number [tex]x[/tex] - multiply it by -2/3, and subtract 9.
[tex]x \mapsto -\dfrac23 x \mapsto -\dfrac23x - 9 = j(x)[/tex]
The inverse operation undoes this, so given some number [tex]j(x)[/tex], we recover [tex]x[/tex] by adding the additive inverse of -9, or 9, then multiplying the whole result by the multiplicative inverse of -2/3, or -3/2.
[tex]j(x) \mapsto j(x) + 9 \mapsto -\dfrac32(j(x) + 9) \mapsto -\dfrac32 j(x) - \dfrac{27}2 = x[/tex]
You seem to be using some version of this method. You made the mistake of thinking the "multiply by -3/2" step applies only to [tex]j(x)[/tex], and not [tex]j(x) + 9[/tex].
