10.86. y = (2 arctg x - x) / (3 arcctg x)

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

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

Substitute and simplify: $$y' = \frac{(\frac{2}{1+x^2} - 1)(3 arcctg x) - (2 arctg x - x)(\frac{-3}{1+x^2})}{9 (arcctg x)^2} = \frac{\frac{2-(1+x^2)}{1+x^2}(3 arcctg x) + \frac{3(2 arctg x - x)}{1+x^2}}{9 (arcctg x)^2} = \frac{(1-x^2)3 arcctg x + 3(2 arctg x - x)}{9(1+x^2)(arcctg x)^2} = \frac{(1-x^2)arcctg x + 2 arctg x - x}{3(1+x^2)(arcctg x)^2}$$.