Answer
$$\frac{dy}{dx}=\frac{68x(1+x^2)^{16}}{(1-x^2)^{18}}$$
Work Step by Step
$$\frac{dy}{dx}=\Big(\Big(\frac{1+x^2}{1-x^2}\Big)^{17}\Big)'=17\Big(\frac{1+x^2}{1-x^2}\Big)^{16}\Big(\frac{1+x^2}{1-x^2}\Big)'=
17\Big(\frac{1+x^2}{1-x^2}\Big)^{16}\frac{(1+x^2)'(1-x^2)-(1+x^2)(1-x^2)'}{(1-x^2)^2}=
17\Big(\frac{1+x^2}{1-x^2}\Big)^{16}\frac{2x(1-x^2)-(1+x^2)\cdot(-2x)}{(1-x^2)^2}=
\frac{68x(1+x^2)^{16}}{(1-x^2)^{18}}$$