Answer
$$ y'= \frac{(1+x^2)\sin(1/x)-2x^3\cos(1/x)}{x^2(1+x^2)^2}.$$
Work Step by Step
Since $ y=\frac{\cos(1/x)}{1+x^2}$, by using the chain rule: $(f(g(x)))^{\prime}=f^{\prime}(g(x)) g^{\prime}(x)$ and by using the quotient rule; $(u/v)'=\frac{vu'-uv'}{v^2}$ and recalling that $(\cos x)'=-\sin x $, the derivative $ y'$ is given by
$$ y'=\frac{-(1+x^2)\sin(1/x) (1/x)'-2x\cos(1/x)}{(1+x^2)^2}
\\=\frac{-(-1/x^2)(1+x^2)\sin(1/x)-2x\cos(1/x)}{(1+x^2)^2}\\ =\frac{(1+x^2)\sin(1/x)-2x^3\cos(1/x)}{x^2(1+x^2)^2}.$$
