Simple question about superposition-theorem
In your calculation the problem is with \$V_2\$. You have a voltage divider, so it should be

$$V_2=3\cdot\frac{9}{6+9}=\frac{9}{5}$$

So the total voltage \$V\$ becomes

$$V=V_1+V_2=\frac{36+9}{5}=9$$

And \$i_x\$ is given by

$$i_x=\frac{V}{9}=1$$

as it should be.

EDIT:

To make it even a bit clearer: you can also directly compute the contributions of the two sources to the current \$i_x\$. Let \$i_{x,1}\$ be the contribution of the current source:

$$i_{x,1}=2\cdot\frac{6\cdot 9}{6+9}\cdot 9=\frac{12}{15}$$

The contribution of the voltage source is \$i_{x,2}\$:

$$i_{x,2}=\frac{3}{6+9}=\frac{3}{15}$$

So the total current is

$$i_x=i_{x,1}+i_{x,2}=\frac{12}{15}+\frac{3}{15}=\frac{15}{15}=1$$

which is of course the same result as obtained above.