If you're using 2:1 and 4:1 multiplexors, your solution yields 10 of the 4:1 gates.
I have a solution that uses four 4:1 and four 2:1 - but it's a bit complicated. In the same way that you can tell whether a number is divisible by 9 by adding up the decimal digits, you can tell is a number is divisible by 3 by adding up the radix four digits. Radix four is binary in groups of 2 bits. So, if A=x1x2 and B=x3x4, the C=A+B (addition, not oring) would give you the total sum of the base 4 digits. But for optimization you don't completely calculate C. With 2 gate (one 4:1 and one 2:1) you can add two bits and generate a carry. So you add D=x1+x3, E=x2+x4, F=D+E and you've used 6 gates.
If you want, you should be able to figure it from there.