Bom dia!
Vamos tentar fazer, então!
\(\left[\dfrac{9+x}{\left(9-x^2\right)^{\dfrac{3}{2}}}\right]'=\dfrac{\left(9+x\right)'\cdot\left(9-x^2\right)^{\dfrac{3}{2}}-\left(9+x\right)\cdot\left[\left(9-x^2\right)^{\dfrac{3}{2}}\right]'}{\left[\left(9-x^2\right)^{\dfrac{3}{2}}\right]^2}
\dfrac{1\cdot\left(9-x^2\right)^{\dfrac{3}{2}}-\left(9+x\right)\cdot\left[\dfrac{3}{2}\cdot\left(9-x^2\right)^{\dfrac{3}{2}-1}\cdot\left(9-x^2\right)'\right]}{\left(9-x^2\right)^3}
\dfrac{\left(9-x^2\right)^{\dfrac{3}{2}}-\left(9+x\right)\cdot\left[\dfrac{3}{2}\cdot\left(9-x^2\right)^{\dfrac{1}{2}}\cdot\left(-2x\right)\right]}{\left(9-x^2\right)^3}
\dfrac{\left(9-x^2\right)^{\dfrac{3}{2}}+3x\cdot\left(9+x\right)\cdot\left(9-x^2\right)^{\dfrac{1}{2}}}{\left(9-x^2\right)^3}
\dfrac{\left(9-x^2\right)^{\dfrac{1}{2}}\cdot\left[9-x^2+3x\cdot\left(9+x\right)\right]}{\left(9-x^2\right)^3}
\dfrac{9-x^2+27x+3x^2}{\left(9-x^2\right)^{3-\dfrac{1}{2}}}
\fbox{\dfrac{2x^2+27x+9}{\left(9-x^2\right)^{\dfrac{5}{2}}}}\)
Ufa!
Espero ter ajudado!