<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] Classical approach:
$$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$$ whereas $u$ and $v$ are functions in $x$. 
So when we have 
$$\frac{2}{x^2+3^3}$$
We have 
$$u(x) = 2$$
$$v(x) = x^2+ 3^3$$
And from that it follows
$$u'(x) = 0$$
$$v'(x) = 2x$$
Using the above formula
$$\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{2}{x^2+3^3}\right) = \frac{0 -2\cdot2x}{(x^2+3^3)^2} = \frac{-4x}{x^4+2\cdot 3^3 x^2 + 3^6} = \frac{-4x}{x^4+54x^2+729}$$
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