If \(y^{\frac{1}{m}}+y^{\frac{-1}{m}}=2 x, x^1 1\), then \(\left(x^2-1\right)\left(\frac{d y}{d x}\right)^2\) is equal to

\(y^{\frac{1}{m}}+y^{-\frac{1}{m}}=2 x \Rightarrow y^{\frac{2}{m}}+y^{-\frac{2}{m}}+2=4 x^2 \ldots .\). (1)
\(\begin{aligned} & \frac{1}{m}\left(y^{\frac{1}{m}-1}-y^{-\frac{1}{m}-1}\right) \frac{d x}{d y}=2 \\ & \Rightarrow\left(y^{\frac{1}{m}}-y^{-\frac{1}{m}}\right)=\frac{2 m y}{\frac{d y}{d x}} \\ & \Rightarrow y^{\frac{2}{m}}+y^{-\frac{2}{m}}-2=\frac{4 m^2 y^2}{\left(\frac{d y}{d x}\right)^2}\ldots(2) \text {[squaring both sides] }\end{aligned}\)
from (1)-(2)
\(\begin{aligned} & 4=4 x^2-\frac{4 m^2 y^2}{\left(\frac{d y}{d x}\right)^2} \Rightarrow 4\left(\frac{d y}{d x}\right)^2=4 x^2\left(\frac{d y}{d x}\right)^2-4 m^2 y^2 \\ & \Rightarrow\left(x^2-1\right)\left(\frac{d y}{d x}\right)^2=m^2 y^2\end{aligned}\)