\[ y'(x)-\frac {\sqrt {y(x)^2-1}}{\sqrt {x^2-1}}=0 \] ✓ Mathematica : cpu = 0.13515 (sec), leaf count = 173
\[\left \{\left \{y(x)\to -\frac {1}{2} e^{-c_1} \sqrt {2 x^2+2 e^{4 c_1} x^2-2 \sqrt {(x-1) (x+1)} x+2 e^{4 c_1} \sqrt {(x-1) (x+1)} x-1+2 e^{2 c_1}-e^{4 c_1}}\right \},\left \{y(x)\to \frac {1}{2} e^{-c_1} \sqrt {2 x^2+2 e^{4 c_1} x^2-2 \sqrt {(x-1) (x+1)} x+2 e^{4 c_1} \sqrt {(x-1) (x+1)} x-1+2 e^{2 c_1}-e^{4 c_1}}\right \}\right \}\] ✓ Maple : cpu = 0.02 (sec), leaf count = 29
\[\ln \left (x +\sqrt {x^{2}-1}\right )-\ln \left (y \relax (x )+\sqrt {y \relax (x )^{2}-1}\right )+c_{1} = 0\]
Hand solution
\begin {equation} y^{\prime }=\pm \frac {\sqrt {y^{2}-1}}{\sqrt {x^{2}-1}}\tag {1} \end {equation}
Separable. For the positive case
\begin {align*} \frac {dy}{dx}\frac {1}{\sqrt {y^{2}-1}} & =\frac {1}{\sqrt {x^{2}-1}}\\ \frac {dy}{\left (y^{2}-1\right ) ^{\frac {1}{2}}} & =\frac {dx}{\left ( x^{2}-1\right ) ^{\frac {1}{2}}} \end {align*}
Integrating
\[ \int \frac {dy}{\left (y^{2}-1\right ) ^{\frac {1}{2}}}=\int \frac {dx}{\left ( x^{2}-1\right ) ^{\frac {1}{2}}}+C \]
But \(\int \frac {dy}{\left (y^{2}-1\right ) ^{\frac {1}{2}}}=\tanh ^{-1}\frac {y}{\left (y^{2}-1\right ) ^{\frac {1}{2}}}=\ln \left (y+\sqrt {y^{2}-1}\right ) \), hence
\[ \ln \left (y+\sqrt {y^{2}-1}\right ) =\ln \left (x+\sqrt {x^{2}-1}\right ) +C \]
For the negative case
\begin {align*} \frac {dy}{dx}\frac {1}{\sqrt {y^{2}-1}} & =-\frac {1}{\sqrt {x^{2}-1}}\\ \frac {dy}{\left (y^{2}-1\right ) ^{\frac {1}{2}}} & =-\frac {dx}{\left ( x^{2}-1\right ) ^{\frac {1}{2}}} \end {align*}
Integrating
\[ \int \frac {dy}{\left (y^{2}-1\right ) ^{\frac {1}{2}}}=-\int \frac {dx}{\left ( x^{2}-1\right ) ^{\frac {1}{2}}}+C \]
But \(\int \frac {dy}{\left (y^{2}-1\right ) ^{\frac {1}{2}}}=\tanh ^{-1}\frac {y}{\left (y^{2}-1\right ) ^{\frac {1}{2}}}=\ln \left (y+\sqrt {y^{2}-1}\right ) \), hence
\[ \ln \left (y+\sqrt {y^{2}-1}\right ) =-\ln \left (x+\sqrt {x^{2}-1}\right ) +C \]
Therefore
\[ \ln \left (y+\sqrt {y^{2}-1}\right ) =\pm \ln \left (x+\sqrt {x^{2}-1}\right ) +C \]