10.84. y = arctg x / x^3

Use the quotient rule: $$y' = \frac{(arctg x)' x^3 - arctg x (x^3)'}{(x^3)^2}$$

Calculate the derivatives: $$(arctg x)' = \frac{1}{1+x^2}$$ and $$(x^3)' = 3x^2$$.

Substitute and simplify: $$y' = \frac{\frac{1}{1+x^2} x^3 - arctg x (3x^2)}{x^6} = \frac{x^3 - 3x^2(1+x^2)arctg x}{(1+x^2)x^6} = \frac{1 - 3(1+x^2)arctg x}{(1+x^2)x^3}$$.