<s>[INST] How to differentiate this fraction $\frac{2}{x^2+3^3}$? $\frac{2}{(x^2+3)^3}$.
I have ${dy}/{dx}$ x 2 x ${x^2+3^3}$ - 2 x ${dy}/{dx}$  x ${x^2+3^3}$ over $({x^2+3)^6}$
And then simplifying to $-12x^5 + 36x^2$  over $({x^2+3)^6}$
I'm not sure if this is right. 
 [/INST] Hint. One may apply
$$
\left( \frac1{f}\right)'=-\frac{f'}{f^2} \tag1
$$ with
$$
f(x)=x^2+3^3.
$$
Find $f'(x)$ then use $(1)$ and conclude with $\dfrac2{x^2+3^3}=2 \times \dfrac1{x^2+3^3}=2 \times \dfrac1{f}$.
</s>