$\begin{align}  & y={{\left( \frac{1+{{x}^{2}}}{1-{{x}^{2}}} \right)}^{\tfrac{3}{2}}} \\ & \text{Take the logarithm of both sides} \\ & \log y=\log {{\left( \frac{1+{{x}^{2}}}{1-{{x}^{2}}} \right)}^{\tfrac{3}{2}}} \\ & \log y=\frac{3}{2}\log \left( \frac{1+{{x}^{2}}}{1-{{x}^{2}}} \right) \\ & \log y=\frac{3}{2}\left( \log (1+{{x}^{2}})-\log (1-{{x}^{2}}) \right) \\ & \text{Differentiate with respect to }x \\ & \frac{1}{y}\frac{dy}{dx}=\frac{3}{2}\left( \frac{2x}{1+{{x}^{2}}}+\frac{2x}{1-{{x}^{2}}} \right) \\ & \frac{1}{y}\frac{dy}{dx}=\frac{3}{2}\left[ \frac{2x(1-{{x}^{2}})+2x(1+{{x}^{2}})}{(1+{{x}^{2}})(1-{{x}^{2}})} \right] \\ & \frac{1}{y}\frac{dy}{dx}=\frac{3}{2}(2x)\left[ \frac{1-{{x}^{2}}+1+{{x}^{2}}}{(1+{{x}^{2}})(1-{{x}^{2}})} \right]= \\ & \frac{1}{y}\frac{dy}{dx}=\frac{6x}{(1+{{x}^{2}})(1-{{x}^{2}})} \\ & \frac{dy}{dx}=\frac{6xy}{(1+{{x}^{2}})(1-{{x}^{2}})}=\frac{6x}{(1+{{x}^{2}})(1-{{x}^{2}})}\times {{\left( \frac{1+{{x}^{2}}}{1-{{x}^{2}}} \right)}^{\tfrac{3}{2}}} \\ & \frac{dy}{dx}=\frac{6x{{(1+{{x}^{2}})}^{\tfrac{1}{2}}}}{{{(1-x)}^{\tfrac{5}{2}}}} \\\end{align}$