Answer
$\frac{\frac{1}{3}x^{2}+3}{(x^{2}+3)^{4/3}}$
Work Step by Step
Factor out $x^2+3$:
$(x^{2}+3)^{-1/3}-\displaystyle \frac{2}{3}x^{2}(x^{2}+3)^{-4/3}=(x^{2}+3)^{-4/3}[(x^{2}+3)-\frac{2}{3}x^{2}]=(x^{2}+3)^{-4/3}(\frac{1}{3}x^{2}+3)=\frac{\frac{1}{3}x^{2}+3}{(x^{2}+3)^{4/3}}$
