This will require the chain rule and using logarithmic differentiation
$\displaystyle y = \frac{(x^2+2)^{\frac{3}{2}}(x^2+9)^{\frac{4}{9}}}{ (x^3+6x)^{\frac{4}{11}}}$
$\displaystyle ln(y) = ln \left(\frac{(x^2+2)^{\frac{3}{2}}(x^2+9)^{\frac{4} {9}}}{(x^3+6x)^{\frac{4}{11}}}\right)$
Since $\displaystyle ln \left(\frac{ab}{c}\right) = ln(a)+ln(b)-ln(c)$ we get
$\displaystyle ln(y) = ln \left[(x^2+2)^{\frac{3}{2}}\right] + ln \left[(x^2+9)^{\frac{4}{9}}\right] - ln \left[{(x^3+6x)^{\frac{4}{11}}}\right]$
Use the chain rule for each of them:
edit: oops no it isn't *edits*
edit2: that's better
Spoiler:
$\displaystyle y' = \left(\frac{(x^2+2)^{\frac{3}{2}}(x^2+9)^{\frac{4} {9}}}{(x^3+6x)^{\frac{4}{11}}}\right) \cdot \left(\frac{3x}{(x^2+2)^{\frac{3}{2}}} + \frac{8x}{9(x^2+9)^{\frac{4}{9}}} - \frac{12x}{11(x^3+6x)^{\frac{4}{11}}}\right)$
I can see no difference between your answer and e^(i*pi)'s, so I would conclude that this is correct.
EDIT: Ahh I see it now. e^(i*pi) did make a slight mistake in the last fraction:
$\displaystyle
y' = \left(\frac{(x^2+2)^{\frac{3}{2}}(x^2+9)^{\frac{4} {9}}}{(x^3+6x)^{\frac{4}{11}}}\right) \cdot \left(\frac{3x}{(x^2+2)^{\frac{3}{2}}} + \frac{8x}{9(x^2+9)^{\frac{4}{9}}} - \frac{12x}{11(x^3+6x)^{\frac{4}{11}}}\right)
$
is supposed to be:
$\displaystyle
y' = \left(\frac{(x^2+2)^{\frac{3}{2}}(x^2+9)^{\frac{4} {9}}}{(x^3+6x)^{\frac{4}{11}}}\right) \cdot \left(\frac{3x}{(x^2+2)^{\frac{3}{2}}} + \frac{8x}{9(x^2+9)^{\frac{4}{9}}} - \frac{12x^2+24}{11(x^3+6x)^{\frac{4}{11}}}\right)
$